Electrical Machinery

MECHANICS DEPT,

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ELECTRICAL MACHINERY

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ELECTRICAL MACHINERY A PRACTICAL STUDY COURSE ON Installation,

Operation and Maintenance

BY F. A.

ANNETT '/

Associate Editor, Power; Member, American Institute of Electrical Engineers; Associate Member, Association

of Iron

and Steel Electrical Engineers

FIHST EDITION

SECOND IMPRESSION

McGRAW-HILL BOOK COMPANY, INC. NEW YORK: 370 SEVENTH AVENUE LONDON: 648 BOUVERIE 1921

ST., E. C. 4

Library

COPYRIGHT, 1921, BY THE

BOOK COMPANY,

INC.

PREFACE

During the three years from January, 1917, to December, " Power" a series of articles on 1919, there was published in practical electricity under the title, "The Electrical Study Course." For the first eighteen months the author of this book wrote the Electrical Study Course Lessons, and during 1919 edited them.

After

many

requests for these lessons in

book form, he was induced to compile these articles into a volume which this work represents. An effort has been made to include such material as would make the work not only of value as a study on elementary theory of electricity and magnetism, but of practical worth. The first eighteen months of the Electrical Study Course " Lessons have been used almost as they appeared in Power," and also parts of the lessons during the following eighteen months. In addition to this there has been incorporated a number of articles by the author on electrical-machinery operation and connections and kindred subjects, which have

"

appeared in Power" during the last seven or eight years. In the author's fifteen years' experience in teaching industrial electricity, directing men on the job and in editorial work, it has been found that there are many features in the study of electricity that a large percentage of home students and those taking short courses find to be stumbling blocks. They get wrong conceptions of the theory or find it difficult to get a clear understanding of certain elementary laws. In the preparation of this material an effort has been made to present the subjects in such a way as to guide the student over these pitfalls and to have all the problems deal with things that the practical f

man comes

^

1

1

U

i

'

in contact with in his

PREFACE

vi

daily work, so as to show how theory fits into practice. Although there are available a number of good books on elementary electricity and magnetism, the demand for the Study Course Lessons in book form has been interpreted as a justification for this volume, and it is presented with the

hope that it has a real field of usefulness. The author takes this opportunity to express his very deep personal appreciation of Mr. Fred R. Low, Editor in Chief

"Power," for his constructive suggestions and inspiration, which has made this book possible. He also desires to acknowledge with thanks the work of Mr. W. A. Miller, of

who wrote last

the Electrical Study Course Lessons during the

eighteen months they were published. F. A.

New York

City,

March, 1921.

ANNETT

CONTENTS PAGE

v

PREFACE

CHAPTER Electricity.

Electricity

1

Sound Analogy. Heat Analogy. Similar to Heat, Sound and Light.

ing System.

How

to

Light Analogy. Electrical Generat-

in a Wire. What Magnetic Effects of an Magnetic Field Varies with the Current. Causes

Electric Pressure

Happens when Current Flows Electric Current.

vi

I

FUNDAMENTAL PRINCIPLES

INTRODUCTION

to

is

Varied

in a Wire.

Changing Direction of Current.

CHAPTER

II

ELEMENTARY MAGNETISM of

the

10 to 28

Property of

Magnets. Artificial Magnets. The Earth a Permanent Magnets. Properties of Magnets. Magnet. Poles of a Magnet. Distribution of Magnetic Field. Lines of Force Conventional. Unit Magnetic Pole. How Magnets Attract Pieces of Iron. Why Iron can be Magnetized. Theory of Magnetism. How Magnets are Demagnetized. Electricity and Magnetism. Direction of Magnetic Field About a Conductor. Relation Between Direction of Current and Magnetic Field. Determining Direction of Magnetic Field About a Conductor. Magnet Field Around a Number of Conductors. Properties of Discovery

Electromagnets.

CHAPTER

III

FUNDAMENTALS OF THE ELECTRIC CIRCUIT

29 to 48 Production of Voltage. Electric Pressure and Temperature. Unit of Electric Current. Electrical Resistance. No Perfect Insulator of Electricity. of Pressure.

Pump Analogy of Electric Circuit. Measurement Measurement of Current Flow. How to Increase

Pump Analogy of Voltage Sources in Series. How Current Voltage. is Increased. Pump Analogy of Current Sources in Parallel. Series-Parallel

and

Parallel-Series Connections. vii

CONTENTS

Viii

CHAPTER

IV PAGE

OHM'S LAW, SERIES AND PARALLEL CIRCUITS Three Electric Units. Volts,

Unit of Electric Current.

Amperes and Ohms. Factors

Generator.

Electric

Ohm's

Current

49 to 72

Law

How that

Problems.

Relation Between

Remember Ohm's Law.

to

Influence

What

Flow

of

Constitutes

Electric Electrical

Flow of Electric Current Compared Series Circuits. Water Flowing Through a Pipe. Voltage Drop Along a Circuit The Value of the Volts is Similar to Pressure Drop in a Pipe Line. Across Each Section of Resistance in Any Circuit May be Found by Ohm's Law. Parallel Circuits. The Total Resistance of Any Resistance. to

Circuit rent.

is

Parallel.

Joint

Equal to the Total Voltage Divided by the Total CurTwo or More Unequal Resistances in Resistance and Conductance. Formula for Finding

Joint Resistance of Resistance.

Two Methods

Hydraulic

Analogy of a

Parallel

Circuit.

Working Out Joint Resistance Problems. Volts Drop Across a Circuit. Determining Value of Current by Volts Drop and Resistance of Section of a Circuit. of

CHAPTER V 73 to 81 ENERGY, WORK AND POWER Energy the Ability of Doing Work. Steam Under Pressure Foot-Pound Unit of Work. Defining Unit of Possesses Energy. Power. Definition of Energy, Work and Power. Horse-power Defined by James Watt. When Energy is Expended Heat is British Thermal Produced. Mechanical Equivalent of Heat. Power of an Electric Circuit. Heat Equivalent of an ElecUnit. tric Current.

CHAPTER

VI

82 to 92 COMPLEX CIRCUITS AND EFFECTS OF INTERNAL RESISTANCE Circuits in Which Elements are Grouped in Series and Parallel. Internal Resistance. How to Compute Value of Current when Effect of Internal Resistance is Considered. Volts Drop in a Source of Electromotive Force. Volts Drop Through Resistance of Armature.

Effects of Increasing the

CHAPTER ELECTRICAL INSTRUMENTS Principles on which

Load on

Batteries.

VII

93 to 106 Instruments are Constructed. Movement for Modern-Type Instrument. Instrument Connected as Voltmeter. Turning Effort Proportional to the Current in the Electrical

CONTENTS

ix

PAGE Coil.

Not Voltage that Causes the

Coil to

Move, but Current.

Instrument Connected to Measure Current. Instrument to Measure Both Volts and Amperes. Direct-Reading Wattmeter. Watt-Hour Meter.

CHAPTER

VIII

METHODS OF MEASURING RESISTANCE Method.

Voltmeter-and-Ammeter

107 to 116

Known-Resistance

Method.

Measuring High Resistance. Measuring Insulation Resistance.; Bridge. Hydraulic Analogy of a Wheatstone Condition to Give Balanced Pressure. Bridge.

The Wheatstone

CHAPTER IX 117 to 132 CALCULATING SIZE OF CONDUCTORS Wire Gage for Round Wire. Circular Mil as a Unit of Measurement. Some Features of the Wire Table. Wire-Number Sizes. Cross-Section and Diameter of Round Conductors. Wire Table Worked Out from No. 10 Wire. Allowable Carrying Capacities of

Copper Wire.

Allowable Voltage Drop.

Economical Size of

Voltage Regulation of Different Circuits. CalculatDerivation of Circular-Mil Formula. ing Size of Conductors. Resistance per Foot of Conductor. Conductors.

CHAPTER X FUNDAMENTAL PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY.

133 to Electromagnetic Induction. Rule for Determining Direction of Electromotive Force. Lines of Magnetic Force. Unit Electromotive Force. Fundamental Principle of the Electric Motor. Direction Conductor Will

Move

in a

.

.

!

Magnetic "Field Depends upon

Counter-Electromotive Force. How an Electric Generator is Loaded. Simplest Form of Electric Machine. Factors Governing Value of the Voltage. Current Reverses in External Circuit. Electromotive Force or Current Curve. Machine with Ring-Type Armature.

the

Direction

of

the

Current.

CHAPTER XI DIRECT-CURRENT MACHINERY CONSTRUCTION Types of Armature. Objections to Ring-Type

152 to 169 Armatures.

Drum-Type Armatures. Effects of Eddy Currents. Eliminating Eddy Currents. Objection to Smooth-Core Armatures. Field Magnets.

Early

Type Machine. the Commutator.

Types Materials

of

Field

Used

in

Magnets. Field

Commutator Construction.

Consequent-Pole Function of

Poles.

CONTENTS

x

CHAPTER

XII

PAGE 170 to 181 INDUCTANCE AND COMMUTATION Magnetic Field About a Conductor. Voltage Induced by DecreasSelf-Inductance. Mutual Inductance. Neutral ing Current. Point on Commutator. How Voltage in Armature is Maintained Constant. Effects of Self-Induction on Commutation. Commutation Period Short.

CHAPTER

XIII

TYPES OF DIRECT-CURRENT GENERATORS

182 to 202 Exciting Field Coils of Direct-Current Generators. How Voltage Lines of Force in Field Poles Does is Built up in the Armature.

Not Increase Coils.

Proportion to the Current Flowing in the Field and Armature Rotation. Reversing Armature. Series-Connected Generator. Ampere-

in

Field-Coil Connections

Rotation of Turns on Field Poles. Load on a Series Generator. Voltage Control on Shunt Generator. Diagrams of Direct-Current Generators. Effects of Loading a Shunt Generator on Voltage Regulation. How Voltage May be Maintained Constant. Compound-

Wound Wound

Generator.

Crossing

Field

Connections

of

Compound-

Generator.

CHAPTER XIV DIRECT-CURRENT GENERATOR CHARACTERISTICS

203 to 214

Voltage Curves of Shunt Generator. Volts Drop in Armature with Fields Separately Excited. Voltage Curve of Series Generator. External Characteristic Curves. Voltage Curve of Compound Generator. Series Winding to Compensate for Volts Drop in Armature. Voltage at Terminals of Compound Generator.

Proper Amount of Compounding. Compounding Shunt. Shunt and Long-Shunt Connections.

Short-

CHAPTER XV LOSSES IN DIRECT-CURRENT MACHINERY

215 to 221 Loss in Transformation of Energy. Friction Losses. Current to Excite the Field Coils. Copper Losses in the Armature. EddyCurrent Losses. Hysteresis Losses. Total Losses in an Electric

Machine.

CHAPTER XVI DIRECT-CURRENT MOTORS Fundamental Principle Brush

Position.

222 to 232 of Electric

Another

Theory

Motor. of

Effects

Motor

of

Changing Both Operation.

CONTENTS

xi

PAGE

Motor and Generator Action Present

in

all

Dynamo-Electric Motor Operates

Machines. Counter-Electromotive Force. How Loaded. How Speed of Motor may be Adjusted.

When

CHAPTER XVII TYPES OF DiRECT-CuKRENT MOTORS

233 to 246 Five Types of Motors. Torque of Series Motor Varies as the Square of the Current. Shunt-Type Motors. Torque of Shunt Motor Varies as the Armature Current. Compound-Type Motors. Effects of Armature Reaction. Commutating-Pole Machines. Number of Interpoles Used. Motors with Compensating Windings.

CHAPTER STARTING RHEOSTATS TORS

XVIII

AND CONNECTIONS TO DIRECT-CURRENT MO247 to 268

Action Taking Place when Motor is Started. Curve. Starting Boxes. Connecting Up a Starting-Current Series Motor. Connecting Up Shunt Motors. Mistakes in Making Starting Resistance.

Connections.

Checking Connections.

Connecting

Up Compound

Testing Polarity of Series and Shunt-Field Windings. Reversing Direction of Rotation. Connecting Up Interpole

Motor.

Machines.

THREE- WIRE SYSTEMS AND

CHAPTER XIX How THEY MAY BE

Circuits. Direct-Current Voltages for Balancer Sets for Three-Wire Systems.

OBTAINED

269 to 289

Three- Wire System. How Balancer Set Main-

Systems Voltage Balanced.; Compound-Wound Three-Wire Generators. Power Output of Three-

tains Three-Wire

Balancer Sets.

Wire Generator.

CHAPTER XX DIRECT-CURRENT GENERATORS IN PARALLEL

290 to 306 Reasons for Operating Generators in Parallel. Effects of Generator Characteristics on Parallel Operation. Compound Generators in Parallel. Equalizer Connection. Putting Generators into Service. Parallel Operation of Three- Wire Generators.

CHAPTER XXI FUNDAMENTAL PRINCIPLES OF ALTERNATING CURRENT

307 to 321

Dynamo-Electric Machines Fundamentally Generate Alternating Distance Traveled and Rate of Motion. Principles of Current. the Dynamo-Electric Machines.

Alternating Electromotive Force.

CONTENTS

xii

PAGE Electrical

Degrees.

A Number

Series Generating Voltage.

of

Conductors Connected

in

Relation Between Cycles, Poles and

Speed.

CHAPTER XXII MEASUREMENT AND ADDITION OF ALTERNATING VOLTAGES AND CURRENT

322 to 332 Alternating-Current Circuits. Voltmeter Readings. Addition of Voltages and Currents.

Instruments

Ammeter

for

and

CHAPTER XXIII TWO-PHASE AND THREE-PHASE CIRCUITS

333 to 349 Systems. Three-Wire Two-Phase Systems. ThreePhase Systems. Three-Phase Machine with Ring Armature. Delta and Star Connections.

Two-Phase

CHAPTER XXIV OPERATING ALTERNATORS IN PARALLEL

350 to 362 Condition for Parallel Operations. Adjusting Load on Two or More Alternators. Putting an Alternator into Service. Other

Methods

Electro-Dynamometer Type of SynSynchronoscope Connections.

of Synchronizing.

chronoscope.

CHAPTER XXV KILOWATTS, KILOVOLT- AMPERES AND POWER FACTOR

How Current Builds Up in a

363 to 375 Lenz's Law.

Direct-Current Circuit.

Lags Behind Voltage. Pump Analogy of Alternating Apparent Watts and Useful Watts. A Wattmeter Reads True Watts. Power in an Alternating-Current Circuit. Current

Current.

Effects of Capacity.

Power

of

Polyphase Circuits.

CHAPTER XXVI POTENTIAL AND CURRENT TRANSFORMERS Types

of

Potential

Transformers.

Principles

of

376 to 385 Transformer

Core-Type and Shell-Type Transformers. Relation Primary and Secondary Volts. Grouping of Transformers.

Operation. of

Current Transformers.

CHAPTER XXVII ALTERNATING-CURRENT MOTORS

386 to 401

Parts of an Induction Motor.

Revolving Magnetic Field. tion Between Poles, Frequency and Speed. How Rotation

Relais

Pro-

CONTENTS

xiii

PAGE duced.

Rotor Must

Run

Slower than the Magnetic Field.

The

Induction Generator.

CHAPTER XXVIII STARTING POLYPHASE MOTORS

Wound-Rotor

Induction

402 to 420 Motors.

Types

of

Rotors.

Wound-

Rotor Motors that Start Automatically. How to Reverse the Direction of Rotation. Comparison of Squirrel-Cage and WoundAuto-Transformers Rotor Motors. How Operate. Starting

Compensators for Polyphase Motors. Compensator. INDEX..

Polyphase Starting

Motors.

Two

or

Rheostats

for

More Motors from

Starting the Same

.

421

ELECTRICAL MACHINERY CHAPTER

I

INTRODUCTION FUNDAMENTAL PRINCIPLES The object of this introduction .is to give a Electricity. clear conception of the electric current and its fundamental actions as used in commercial electrical machinery. It will be confined to a pictorial description rather than a mathe-

matical analysis or study of the laws of electricity, because will be easier for the student to understand the operation

it

of the machinery that will be described later if a practical view is obtained of what is happening on the inside rather than to try to understand the results of laws or equations. Later, these fundamental principles will be referred to a great many times to get an insight of the actual workings of

the different electrical machines, therefore reader to get them firmly fixed in his mind.

it

will

pay the

Electricity is only one of several peculiar happenings in nature which are very common to us, but which are more or less mysterious. Sound, heat and light are in many respects similar in action to electricity, and on account of their being more easily understood it will be advantageous to compare a few of their actions. In each case there is a generating, a

transmitting and a receiving apparatus. In each the product is invisible as it travels along the transmission system consequently, what has happened must be judged from the ;

effects

produced.

Sound Analogy.

If a pencil is tapped on the desk, the contact generates a sound. This sound is transmitted in 1

ELECTRICAL MACHINERY

2

every direction along many paths from the point of contact. One path is through the air to the ear. The air is the transmission and the ear is the receiving apparatus. If an easier path having less resistance to the is substituted, the listener will get

along

sound as it travels more sound at the

For instance, if the ear is held to a railroad someone hits the rail a blow a half-mile away, the It could not be heard Jsp-iind: can be heard through the rail. through the air. The rail is the better conductor. It will also be noted that the sound cannot be seen as it travels along the rail or through the air. If the pencil is tapped on the desk with a greater speed, more sound will be transmitted to the ear than if tapped with a slow motion. There are two ways, therefore, of getting more sound improving the conductor by using one of less resistance, and increasing otjier end,

track; ,and

the speed of the generator.

This

is

similar to the conditions

an electrical generating system. Heat Analogy. If one end of an iron rod is held against an emery wheel and the other in the hand, the emery cutting across the iron will generate heat, and the heat will flow along the rod to the hand. Thus we have a heat-generating system a generating, a transmitting and a receiving apThe emery wheel cutting across the iron is the paratus. generator, the rod is the transmission and the hand is the

in

Here, too, the heat transmitted will depend upon the speed of cutting and the nature of the conductor. If a wooden rod is substituted, very much less heat will flow. If receiver.

a

vacuum

is

inserted in the line of transmission, practically it that is, it is an insulator to the

no heat would flow across flow of heat.

It will

;

be noticed in heat transmission also that

nothing can be seen moving along the line of flow. Its effects must be dealt with, and it will be found that electricity is a close relative of heat and similar to it in some respects.

Light Analogy.

Light also has

many

characteristics that

are similar to electricity. When the filament of an incandescent lamp is heated to a certain temperature, it begins to

INTRODUCTION FUNDAMENTAL PRINCIPLES send out eye.

light.

The

It travels at

3

light is transmitted along a line to the much faster than heat

a very rapid rate,

Electricity travels at even a faster speed than light. light travels along lines in space, it is also invisible to the eye. It is only when the end of one of these lines

or sound.

As

reaches the eye, either direct or reflected, that the light is detected by the eye. This is clearly shown by the moon. At night the space around the moon is full of light rays, or lines of light from the sun, but it is only those lines that strike the

moon and light.

are reflected to the eye that give the sensation of The other lines of transmission of light are invisible

and the passage of

light along

them cannot be

seen.

Electricity Electricity Similar to Heat, Sound and Light. is in several ways similar to these phenomena of heat, sound and light. It is invisible and for that reason we must use our

imagination and be governed by its effects. It is not known exactly what it is, but that need not confuse the student, because there are other very common things in the same class. When the rod was held on the emery wheel, heat

We

do not know exactly that moves along the rod, but we know its effects. Light passing from the lamp must send something along the line to the eye. It is said to be wave motion and that no passed along the rod to the hand.

what

it is

material substance actually passes. To make this clearer suppose a wire on a pole line is struck with' a pair of pliers.

wave to travel along the wire to the next was the wave that went to the next pole and gave pole. it a sensation, no material substance actually passed along This will cause a It

That is the way light travels through space, but There is some sort of a at a speed a million times faster. vibration at the generator of light a wave is sent along a line in space and finally reaches the eye and gives it a sensa-

the wire.

tion similar to that which the pole received. Electricity is a similar action of the internal substance of the wire. Nothing

actually passes, but as with light, heat and sound, we say flows along the wire, which is the transmitting medium. Electricity must, therefore, be studied

from

it

its effects,

ELECTRICAL MACHINERY

4

or rather, the effects of electricity must be studied, because they are, after all, what we deal with in all electrical

machinery.

As

example of an

in the case of heat generating, take a simple electrical generating system and study the

effect in each unit apart. By understanding the fundamental actions that occur in each part, the more complicated electrical machinery, which is simply an enlargement of

these actions, will be simplified. Electrical Generating- System. If a magnet is used in place of an emery wheel and a copper wire in place of the rod, Fig.

we have an 1.

FIG.

A

1.

electrical generating system, as

magnet

Illustrates

(for our purpose)

is

shown

in

a piece of steel that

fundamental principle of

electric generator.

gives out magnetic lines of force. If this magnet is rotated so that the end passes across the copper wire, not quite touching it, and the magnetic lines coming out of the end of

the magnet are cut across by the wire, electricity will be generated in the loop. The emery, by actual contact with

when

cutting across it, generates heat in the rod. The magnetic lines, by actual contact with the copper wire when cutting across the latter, generate electricity in the

the rod

conductor.

In the

first

case

it is

said to be the friction of

the particles of emery on the rod that generates the heat. In the second case it may be said that the friction of the

INTRODUCTION FUNDAMENTAL PRINCIPLES

5

magnetic lines cutting across the wire produces electricity. We do not use the word friction, however, because that is used in connection with heat generation. We say in electricity generation it is the induction of the magnetic lines cutting across the wire, but it means practically the same thing. This, therefore, is the all-important principle to

ber

magnetic

lines

generate electricity.

across

cutting This action

is

a

closed

remem-

wire

loop use of in gen-

made

motors and in fact, all commercial apparatus except heating or chemical appliances. The magnetic lines are made to cut across wires by several methods, but fundamentally it should be remembered that the cutting erators, transformers,

action of the magnetic lines across the wire generates the electricity.

How

Electric Pressure Is Varied in a Wire.

The reader

has probably noticed one thing in the figure for electricity generation that is different from that of heat generation a return path is shown from the receiving apparatus. This is necessary in electricity or the current will not flow. The cutting of the magnetic lines across the wire generates an

tending to send the current along the but no current will flow until the complete path is wire, made. This electrical pressure is called voltage. If the magnetic lines cut across the wire at a very slow rate, there is not as much friction or induction, and the pressure, or electrical pressure

voltage,

is

low.

If the speed of cutting the

magnetic

increased, the pressure or voltage is higher. directly proportional to the speed of cutting. If the

number

will not be as

lines is

The pressure

is

of the magnetic lines is decreased there friction, or induction as it is called,

much

hence, the pressure or voltage will be less; also, if the number of lines cut is increased, it will proportionally increase

the voltage. It may therefore be seen

how a pressure, or voltage, is generated in the wire, which tends to send an electric current along the wire. The amount of current that will flow

ELECTRICAL MACHINERY

6

due to this pressure depends upon the resistance in its path. The amount of sound that came from the blow on the rail

depended upon the resistance to the flow of the sound. Very little came through the air most of it came through the rail. The rail had less resistance to the sound. The electric current will follow the path of least resistance in its endeavor to return to the generating point, and the amount of current, called amperes, that flows around this path will be governed ;

by the resistance of the path for a given voltage. If the voltage is raised by speeding up the cutting or by increasing the number of lines, there will be more amperes flowing for a given resistance in the path. Therefore, the current depends the resistance of the path and the pressure, or voltage, behind it. The voltage depends upon two things

upon two things

the speed of cutting the magnetic lines and the number of magnetic lines cut. What Happens When Current Plows Through a Wire. also

is now flowing along the wire. Let us consider the things that af e happening in and around the wire while the current is flowing through it. It has already been said that electricity is closely related to heat. It is a peculiar fact

The current

that electricity flowing along a wire produces heat. The amount of heat produced will depend upon the amount of

current flowing and the amount of resistance offered. If practically all the resistance is in one place, that part will get the hottest. Copper wire is used to conduct the current

because

it offers

very

little

resistance to the flow.

When

it

desired to produce heat, as in a flat iron, a section of conductor is placed in the path which offers a high resistance.

is

A

close relative of heat is light. The wire in the electric by forcing an electric current through it,

bulb, if heated

quickly becomes hot enough to send out light. Light waves go out from the wire, due to the heat, and the heat is due to the electric current passing through a high resistance. Besides the heating effect, there is the chemical effect of an electric current, but on account of its very small use in

commercial machinery,

it

will not be

taken up at this time.

INTRODUCTION FUNDAMENTAL PRINCIPLES

7

Magnetic Effect of an Electric Current. There is one the magnetic effect. This is the effect from which all mechanical power from

other effect of the electric current electricity is secured.

As the current

passes along the wire

from the generator it sends out circular lines of magnetism around it. Electricity and magnetism are as closely related There may be heat without light and as heat and light.

magnetism without electricity, but not light without heat, and similarly electric current cannot be had without mag-

Whenever current

is flowing in a wire, there are the conductor, and it is often for magnetic surrounding the sole purpose of getting these magnetic lines that an

netic lines.

lines

Except for heating, the magnetic surrounding the wire are what we are after in all electrical machinery. When the path, or circuit, is completed in the simple generating system, the current flows and the magnetic lines spread out around the wire. When the circuit is opened, electric current is used.

lines

they close in and vanish.

This will, perhaps, be

made

clearer

by watching the hairspring of a watch. Let the shaft of the balance wheel represent the wire and the hairspring the magnetic lines that surround it. Let the tick of the escapement represent the opening and closing of the circuit. On one tick the circuit is closed, the current flows and the magnetic lines spread out around the wire. Another tick, the circuit is broken, the current stops and the magnetic lines close in. If the current is left on and is continuous in the

same

direction, the magnetic lines will spread out to a certain value and remain stationary as long as the current re-

mains the same.

Magnetic

lines

cannot be produced with

pressure or voltage alone on the wire; the circuit must be closed and current flowing.

Magnetic Field Varies with the Current. The number of surrounding the wire will depend upon the amount of

lines

current, or amperes, flowing. If the current in the wire is reduced, the surrounding magnetic lines close in, and if the current is reduced to zero, they do not immediately vanish.

ELECTRICAL MACHINERY

8

but close to zero gradually. If the current is increased, the magnetic lines spread out and more take the place of those spreading out. This spreading out and closing in of the lines is

magnetic

of these lines

is

extremely important, because the motion of in cutting across near-by wires.

made use

This action will be referred to a good

many

times in later

chapters.

Nothing has been said as yet regarding the direction of This is important, because it the direction of the magnetic lines about the wire. ing to the simple generating system, Fig. 1, it will that the ends of the permanent magnet are marked current in the wire.

N

governs Returnbe seen (north)

and S (south) pole. The magnetic lines are supposed to come out from the north pole of the magnet and return to the steel through the south pole. There will be a different effect

produced in the wire when

it

cuts through the lines of

force from the north pole than when through those from the south pole. One will induce a rvressure tending to send the

current around the loop in one direction, and the other will tend to send the current around in the opposite direction. That is, the direction of the current in the wire will alternate

every time the magnetic lines from a different kind of a pole In other words, an alternating current is produced. Also, if the direction of the cutting is changed from

cut the wire.

upward

to

downward with

the same pole, the direction of the

induced current will be changed. Causes Changing Direction of Current. portant to remember that the

direction

It is

of

the

induced in the wire will alternate from two causes

very imcurrent

one due

changing the direction of the magnetic lines and the other due to changing the direction of the wire across the lines of

to

force.

There

is

no change in the heating

in direction of the current.

which direction the current

due to the change same no matter in There is a change, howeffect

It is the

flows.

As the current starts in one the direction, surrounding magnetic lines start around the ever, in the magnetic effect.

INTRODUCTION FUNDAMENTAL PRINCIPLES

9

wire in a given direction and spread out as the current maximum value; then, as the current decreases to start in the other direction, the magnetic lines

increases to its

close in until the current

becomes

zero.

When

the current

the magnetic lines go around the wire in the opposite direction and spread out as the current increases. To remember this change in direction, imagine a screw going

reverses,

As the screw goes in the direction of the curthe threads will go around in the direction of the rent, lines. As the screw is withdrawn with the current magnetic into a hole.

coming toward the observer, the threads

will go

around

in

opposite direction. This action is continually changing as the current goes in one direction, then in another. As the current changes direction, it does not chop off suddenly and

go in the opposite direction. It decreases gradually to zero. changes direction, increases gradually to its maximum value then decreases to zero again. This increasing and decreasing of the current is accompanied bv an increasing and decreasing of the surrounding magnetic

lines.

CHAPTER

II

ELEMENTARY MAGNETISM Discovery of the Property of Magnets, Before a clear conception can be obtained of electrical machinery, one must first understand the principles of magnetism. Magnets in

some form or other enter into all machines for the generation and utilization of electricity. Consequently, at the beginning of this electrical study course our attention to this subject.

it is

essential that

we

direct

The discovery of the property of magnets is something come down to us from the ancients the exact dates not are known. Magnets were first found in a natural state certain iron oxides were discovered in various parts of the world, notably in Magnesia in Asia Minor, that had the property of attracting small pieces of iron. Fig. 2 shows one

that has

;

;

of these magnets

;

as will be seen,

it

looks very

much

like a

piece of broken stone. Fig. 3 is the same magnet after has been placed in a box of small nails and withdrawn.

The property which the natural magnet possesses attracting small pieces of iron or steel

it

of

called magnetism, and any body possessing this property is called a magnet. Outside of iron and steel there are very few substances that

a

magnet

will attract

;

nickel

is

and cobalt are the

best,

but the

magnetic properties of these metals are very inferior 'compared with those of iron and steel.

when

Magnets. When a piece of iron or steel is under the influence of a natural magnet, it takes on brought the same property; that is, it also becomes a magnet and is said to have been magnetized. The greatest effect is obArtificial

tained by bringing the iron or steel in contact with the One of the remarkable features of this is that the

magnet.

10

ELEMENTARY MAGNETISM

PIG.

FIG.

3.

2.

11

Natural Magnet.

Natural magnet attracting iron

nails.

ELECTRICAL MACHINERY

12

magnet imparts its property to the piece of iron or steel without any apparent loss of its own. Why this is so will be discussed when we consider the theory of magnetism later in this chapter.

A

magnet that

is

made by bringing

iron under the influence of another

a piece of steel or

magnet

is

called an

be divided into

magnets may temporary and permanent magnets. A piece of iron brought under the influence of a magnet becomes magnetized, but when it is removed from this influence, it loses artificial

two

magnet.

Artificial

classes

the property of a magnet almost entirely, therefore, is called a temporary magnet. That is, it remains magnetized only as long as it is under the influence of another magnet.

Permanent Magnets. On the other hand, if a piece of steel is rubbed on a natural magnet, it will become magnetized, and when removed from this influence, it will retain its magnetic properties. Hence we have a permanent magnet. Permanent magnets are usually made in two forms; namely, bar magnets and horseshoe magnets. When made hard

in straight bars, as in Fig. 4, they are called bar magnets,

and when in U-shape, as in Fig. 5, horseshoe magnets. The natural magnet, when suspended from the end of a thread, possesses the property of pointing north and south the same as the needle of a compass. This discovery was used to advantage by the ancients to direct the course of

Hence the natural magnet got the name lodestone (meaning leading stone). Properties of Magnets, If a bar magnet is suspended from a string, as in Fig. 6, it will take a position almost

their ships at sea.

directly north and south. The end that points toward the north pole of the earth is called the north or N pole of the magnet. The end pointing to the south pole of the earth is called the south or S pole of the magnet, as indicated in the figure.

magnet is suspended as in Fig. 7 and the N pole of magnet brought near the N pole of the one that is suspended, it will be found that they will repel each other; If a

a second

ELEMENTARY MAGNETISM that

is,

that

is

the suspended

brought near

the other hand,

if

magnet

it,

will

13

move away from

as indicated

the one

by the arrow-head.

the S pole of the second magnet

is

On

brought

Fia.7

Thread

i

Pole

H

Fra. 9

FIGS. 4 to 10.

Illustrate

X

\ . South j>- N Magnetic" ~~\^."!'

*

-^Geographical

^~C^ North Pole

X^ljg /'

Fia. 10

some of the properties of magnets

N pole of the suspended magnet, as in Fig. 8, it will be found that they will attract each other and the end of the suspended magnet will be drawn to the one held in the hand.

near the

The same effect will also be found at the S pole of the suspended magnet; namely, between like poles there is repul-

ELECTRICAL MACHINERY

14

sion, and between unlike poles there is attraction. From this we may write the first law of magnetism, which is: Like

and unlike poles attract. Carrying this one step farther and holding the second magnet under the one that is suspended, a condition will

poles repel

shown

The suspended magnet will take a 9. of the one held in the hand. that with position parallel will it found that in this case unlike poles be Furthermore, in as the out attract, pointed preceding paragraph. The Earth a Magnet. When the bar magnet was suspended from a string and not influenced by anything else, it pointed approximately north and south, just as it took a position parallel to the second magnet in Fig 9. Therefore we are in a position to. make the deduction that the earth is also a magnet, for it has the same influence upon a magnet as the second magnet had in Fig. 9. This is true. The earth is a magnet, and any effects that may be obtained from a magnet may be obtained from the earth 's magnetism. In Figs. 6 to 9, it was pointed out that unlike poles attract and like poles repel. It was also shown that the end of the magnet that pointed to the north pole of the earth is called exist as

in Fig.

N

pole. Consequently, if unlike poles attract, the magnetic pole of the earth must be opposite to that of the magnet; that is, the earth's magnetic pole located near the

the

geographical north pole is a south magnetic pole and that located near the south geographical pole of the earth, a north

magnetic pole. The true condition

is shown in Fig. 10, which the earth and its The position that a comrepresents poles. in the figure. It will be pass needle will take is indicated

seen that the compass does not point due north and south. In order that ships may be steered by a compass, magnetic charts must be used showing the correction that must be made for the declination of the compass needle.

Poles of a Magnet.

The ends of a magnet are called

its

poles and the distance midway between the poles the equaIt is at the poles that the magnet tor, as shown in Fig. 11. possesses the greatest attraction for pieces of iron, and this

ELEMENTARY MAGNETISM attraction decreases until the center,

15

or equator, of the

reached, where it becomes zero. This attraction magnet is represented in Fig. 12, where a magnet is shown attracting three pieces of iron at its poles. At a distance in from the is

will attract only two, a little nearer the equator it and at the equator none. The same effect

it

pole

will hold but one,

may be obtained if the magnet is placed in iron filings and then withdrawn, as indicated in Fig. 13. Here it is seen that the filings are bunched at the ends of the magnet, but decrease toward the center where it is clean. The space outside of a magnet is occupied by a field of and

influence field.

magnet or a magnetic from the N pole and around through space and enters at the S pole, as is

called the field of the

This field of influence emanates

circulates

indicated in Fig. 14.

It will

be noticed that the greatest

emanate from and enter at the poles, although some of them leave and enter at the sides of the magnet. That some of the magnetism exists along the sides of the magnet would be expected from the results seen in Figs. 12 and 13: there must be magnetism around the sides of the magnet or it would not have attracted the pieces of iron but, as pointed out before, it becomes less as we approach the equator, where it is zero. Distribution of Magnetic Field. There are many ways of

number

of

lines

;

determining the distribution of the magnetism about a magnet.

One of the simplest is to place the magnet under a piece and then sprinkle iron filings upon the latter. By

of glass

gently rapping the glass, the filings will take definite paths, as indicated in Fig. 15. It will be seen that, although the lines are broken, they nevertheless conform very closely to those

shown in Fig. 14. After the filings have been arranged show the direction of the magnetic field, if the glass

so as to is

placed over a sensitized paper, a photograph

of the formation.

a

known by

field of influence

may

be

is

magnetic flux, flux

made

from the poles of

various names, such as magnetic from the magnet or field poles, and but they all mean one and the same thing.

magnet

of force,

The

field,

lines

ELECTRICAL MACHINERY

16

Another method of determining the direction of a magis illustrated in Fig. 16. A magnetized needle is a bottle cork and in a vertical posifloated pushed through

netic field

tion in a vessel containing water. If the dish is placed over a bar magnet with the S pole of the needle over the S pole Equator\

-''

-~

__.

^#rfTlTn UfiWl I

Fia. 16

FIG. 15

FIGS. 11 to 16.

Show

of the magnet,

it

the distribution and direction of a magnetic

will be

field.

found that instead of the cork with

N

the needle floating along the axis of the magnet to the pole, it will take a curved path like that indicated in the figure.

This

is

similar to the path

in Fig. 15. Lines of Force Conventional. field is

expressed as so

many

shown by the iron

filings

The strength of a magnetic

lines of force per square inch

ELEMENTARY MAGNETISM

17

or square centimeter. This naturally brings up the question of what constitutes a line of magnetic force or how these

counted or measured. In the first place, the term force" is only figurative, because in the true sense, or at least as far as we know, such a thing as a definite line does not exist. Therefore, the lines which are generally shown in pictures of magnets are used only to indicate that

lines are 1

'line of

a flow of "something" from the magnet and the direction of the flow.

there the

is

N

to the S pole of

we were to place two fans so that when they were runone would be blowing against the other, the force of the ning air would tend to cause them to separate. If they were suspended from long cords and a spring balance connected between them, the force that one exerted on the other could be measured. Suppose, for example, that the fans tend to separate with a force of 10 oz. If, instead of calling this force ounces or pounds, we called each ounce of repelling force a line of air, it would be the same thing that is done in measuring the strength of a magnetic field only in the magnetic measurements a unit called a dyne is used. This unit is known as the If

;

centimeter-gram-second

(c.g.s.)

unit and

equivalent to

is

part of a pound.

Unit Magnetic Pole.

If the like poles of

two magnets are

placed one centimeter (approximately 14 in.) apart, as shown in Fig. 17, and they repel each other with a force of one ^^

FIG. 17.

-

-

-I,

Shows how

like poles of

magnets

repel.

dyne, the magnetic field is said to be of unit strength. Therefore one line of magnetic force is just what the term

would imply; namely, a

force,

just

as one

fan blowing

ELECTRICAL MACHINERY

18

against the other created a force which tended to separate them.

To express the magnetic unit more accurately, when two magnets of equal strength, of one sq.cm. cross-section each, are placed one centimeter apart in air and repel each other with a force of one dyne, they are said to be of unit strength. In this case the field of the magnet

is

said to be of unit

intensity that is, one line of force per square centimeter and is called a gauss, after Karl Frederick Gauss, a German scientist. So we see there is nothing mysterious about this is it a unit of measurement just as the pound or unit;

ounce.

How

Magnets Attract Pieces of Iron. In a foregoing paragraph in this chapter we found out that a magnet would attract a piece of iron or steel, but had little or no attraction for other substances.

"Why it

is this

so?"

will attract.

This naturally brings up the question, Anything that a magnet can magnetize

When we

say that a magnet will attract a

piece of iron, it is not a true statement. What really happens is that the piece of iron is first made a magnet and then

the magnetism in one attracts the magnetism in the other. In Fig. 18 is shown a bar magnet M. If a piece of iron F is

FIG. 18.

Shows how a magnet

attracts a piece of iron.

become a magnet having N and S be seen that the end of the iron nearest the S pole of the magnet has become an N pole and are passing that some of the lines of force from magnet iron. have the of We piece previously found out through that unlike poles attract, therefore if the end of the iron next

placed as shown,

it

poles, as indicated.

will

It will

M

to the south pole of the

magnet

is

made

a north pole,

it

will

ELEMENTARY MAGNETISM be attracted by the S pole of the magnet. This will not attract other substances because

magnet

19 is

why

it

cannot

a

make magnets out of them. Why Iron Can Be Magnetized.

first

The foregoing brings up another important question, "Why can a piece of iron be magnetized and most other substances not?" The answer found in the theory of magnetism. If a bar magnet broken into several pieces, as shown in Fig. 19, each piece

to this is is

FIG. 19.

will be

Results of breaking bar magnet in a

found

to be a

number of

parts.

magnet with a north and a south

pole,

This experiment can be easily made by maga hack-saw blade and then breaking it into several netizing Each will be found to be a magnet. If the pieces. piece as shown.

parts are placed together again it will be found that the magnetic poles of each piece will disappear at the division

and be present

FIG. 20.

at the

two ends, as shown

in Fig. 20.

There

Results of placing together parts of magnet, Fig. 19.

one conclusion to draw from this fact, and that is, if the magnet is broken up into infinitely small pieces, each piece will be a magnet with a north and a south pole. Theory of Magnetism. There are many theories of magnetism, but the one given preference to-day is based on the It must be remembered that just foregoing hypothesis. is

what magnetism

What

is

a scientific question that is not settled. generally accepted is that each molecule of iron is is, is

ELECTRICAL MACHINERY

20

a magnet in itself. Under natural conditions the molecules take a position so that the poles of one neutralize those of the other, as shown in Fig. 21, just as the poles of the pieces in Fig. 20 neutralize each other when they were brought together. When the pole of a magnet is brought near a piece of iron, it will attract the opposite pole of the

of

magnet

molecules and pull them around in a systematic form, as in Fig. 22, and the lines of force of one molecule will pass into the other

and appear only

Pia. 21

r

end of the piece of iron, magnet did in Fig. 20.

at the

just as the poles of the pieces of

ELEMENTARY MAGNETISM

21

arrangement after they have once taken such a position, and the steel remains magnetized, or in other words, a permanent magnet is obtained.

How

Magnets Are Demagnetized. After steel is once magnetized, it will remain so indefinitely unless subject to some outside influence such as that of another magnet or vibration. If a permanent magnet is hit a few times with a hammer, it will be found to have lost its magnetism. This is caused by the molecules vibrating

was

The

hit.

when

the piece of steel

effect is similar to that of placing grains of

sand on a plane that is slightly inclined. As long as the plane remains motionless, the sand will not move down its surface, the friction between the plane and the sand being prevent the latter from moving. If the plane struck with a hammer and caused to vibrate, the sand sufficient to

is is

loosened up from the surface, or in other words the friction is decreased, and it gradually works down the plane. Similarly, in the permanent magnet when the molecules are caused to vibrate,

hammer

hit with a which decreases

it is

the friction between them, and they gradually pull each other around to a position where their poles neutralize. The

same effect is obtained by heating the magnet. This is due to two reasons: First, the steel loses its hardness, and secondly, heating causes the molecules to vibrate as magnet is struck with a hammer.

In the foregoing

was pointed out that

a

the

magnet im-

properties to a piece of iron without any apparent From what we have found out about the theory of

parted loss.

it

when

its

evident that a magnet does not really impart to a piece of iron, but rather makes manifest property those properties already in the iron; therefore, the magnet is not subject to any action that would cause a loss.

magnetism

it is

its

As pointed out in Chapter I, Electricity and Magnetism. there may be magnetism without electricity, but not electricity without magnetism, just as heat can be produced without

What

light,

but light cannot be produced without heat. is to the other is a scientific ques-

the relation of one

ELECTRICAL MACHINERY

22

is not settled. But that has nothing to do with our study; suffice it to keep the fact in mind that when an electric current flows through a wire, it sets up a magnetic field about it. This is a very important feature in the whole

tion that

study of electricity and magnetism. Direction of Magnetic Field About a Conductor. The directions of the lines of force about a conductor with a current of electricity flowing through it will depend upon the direction of the current. The direction of the current and the corresponding direction of the magnetic flux about a conductor is indicated in Figs. 23 and 24. In these figures

A i 4 //g||\\ V V V

FIG. 24

FIG. 23

FIGS. 23 and 24.

Show

direction of field about

an

electric conductor.

the full-line circle represents the cross-section of the conductor, the dotted circle and arrow-heads the magnetism and its

direction,

and the

cross

and dot in the center

of the

important that the two latter conventional be firmly fixed in the mind, since they are frequently used to represent the direction of current flowing in a conductor. In Fig. 23 the cross represents circles the direction of the current.

It is

current flowing away from the reader. It will be seen that the lines of force are whirling in a clockwise direction. The dot in the center of the circle, Fig. 24, indicates current This is just flowing up through the plane of the paper. opposite to that in Fig. 23, and as will be observed, the direction of the magnetic 5eld about the conductor has is now in a counter-clockwise direction.

reversed, or

Relation Between Direction of Current and Magnetic

These things are so simple that it is hard to remember just what each represents therefore a rule will not be out of place. Referring to the direction of the current in the

Field.

;

ELEMENTARY MAGNETISM

23

conductor, the cross and dot in the center of the circle may be considered to represent an arrow moving in the direction of the current, the cross representing the tail of the and the dot the point. Consequently, if we wish to

arrow

know

which direction of current the cross represents, we may reason like this The cross represents the tail of the arrow, therefore the arrow must be pointing away from the ob:

server for this part to be seen. Hence, the cross represents current flowing away from the reader. Likewise for the dot The arrow must be it represents the point of the arrow. ;

pointing toward the observer for the point to be seen, consequently, the direction of the current which it represents. The next thing to remember is the direction of the lines of force about the conductor.

A

simple rule for this

is

to

grasp the conductor in the right hand with the thumb pointing in the direction of the current, as indicated in Fig. 25 ;

FIG. 25.

Eight -hand rule for telling direction of

field

about an

electric conductor.

the fingers will then point around the conductor in the direction of the lines of force.

The field about a wire carrying an electric current does not take the form of a spiral, but a circle, as shown in the These circles seem to emanate from the center of figures.

and have an increasing diameter as the curThis is illustrated in Fig. 26, a, &, c, and d. At a is represented the effect that would be obtained from a small current flowing in the conductor; at Z>, c, and d, the the conductor

rent increases.

successive development of the magnetic field as the current flowing through the wire increases.

ELECTRICAL MACHINERY

24

Determining Direction of Magnetic Field About a ConThere are several ways of determining that a magnetic field is set up about a conductor carrying an electric ductor.

current.

If a wire is

FIG. 26.

placed through a hole in a piece of glass

Illustrates

how

field

builds up about a conductor as

current increases.

or cardboard, as shown in Fig. 27, and an electric current is caused to flow through the wire, then if iron filings are

sprinkled upon the glass or cardboard, they will be arranged in broken circles, as indicated in the figure. To get the best effect it will probably be necessary to tap the edge of the

FIG. 27.

Showing magnetic

field

about a conductor with iron

filings.

Placing a compass in various positions as shown, always point in a definite direction around the conductor. It will be observed that the compass points in an

surface. it

will

opposite direction on opposite sides of the wire. This fact may be used to determine the direction of an electric current in a conductor. In Fig. 28 is represented a

wire carrying a current in the direction of the arrow; the magnetic field is shown by the dotted circles and arrow-

ELEMENTARY MAGNETISM

25

If a compass is placed on top of the conductor, it will point in the direction of the magnetic field above the wire, likewise if placed underneath; namely, as indicated in the After the direction of the magnetic field has been figure.

heads.

determined by the compass,

FIG. 28.

Shows

if

the wire

direction of compass

is

grasped in the

when placed near an

electric conductor.

right hand- with the fingers pointing around the conductor in the direction that the compass points, the thumb, when

pointing along the conductor, will indicate the direction of the current, as illustrated in the figure .

Magnetic Field Around a Number of Conductors. In Fig. 29 are represented two parallel conductors having an

FIG. 29

FIG. 30

FIGS. 29 and 30.

Direction of magnetic field about two conductors carrying current in the same direction.

electric current flowing in the same direction in each and supposed to be far enough apart so that the magnetic field of one will not have any influence on that of the other. The

direction of the magnetic field between the

two conductors

ELECTRICAL MACHINERY

26

an upward direction on one and downtwo conductors are brought near together, the two fields between them will oppose each other and since the field strength of both is the same, no lines of force will pass between the wires, for be it remembered that will be seen to be in

ward on

the other.

If the

equal opposing forces in magnetism neutralize each other just as equal opposing mechanical forces neutralize each other, but this does not destroy the

The

result

of

the

magnetic field in this two conductors close

case. bringing together is shown in Fig. 30. It will be seen that the whirls about one wire have combined with those of the other, and now the lines circle the two conductors instead of one and

same relative direction as that in Fig. 29. of conductors are placed alongside of one another with an electric current flowing in the same direc-

this field has the If a

number

tion in them, all their magnetic fields will combine and form a field that will encircle the group. This is the effect ob-

tained in a coil of wire connected to a source of electricity. In Fig. 31 is shown the cross-section of a coil of four turns

;

FIG. 32

FIGS. 31 and 32.

Show how

the magnetic field

is set

up

in a solenoid.

the direction of the current through the coil is indicated by the crosses and dots. That is, it flows in at 1 and oiit at 2,

and out at 4, etc. The turns are supposed to be separated far enough so that the flux of one will not affect that of the other. From an examination of the figures it will be in at 3

seen that the direction of the lines of force on the inside of point in the same direction, likewise on the outConsequently if the turns are placed together as in

the coil side.

all

ELEMENTARY MAGNETISM

27-

Fig. 32, the magnetic fields about the different conductors will coalesce into one and surround the whole group of con-

ductors, as shown. That is, we would have a magnetic field emanating at one end of the coil and entering at the other, just as we did with the permanent and natural magnet. In the case of the coil, however, it is called a solenoid, .but it has It will attract all the properties of a permanent magnet.

pieces of iron, and if suspended by a string so that the axis of the magnetic field is in a horizontal position, it will point

north and south just the same as a permanent magnet. Properties of Electromagnets. If the coil is supplied with a soft-iron core, as in Fig. 33, the combination is called

FIG. 33.

Field about an electromagnet.

an electromagnet. Although this device possesses the properties of a permanent magnet, it will remain magnetized only as long as current

is

flowing in the

coil,

and

is

therefore a

temporary magnet.

A

convenient rule for determining the polarity of an when the direction of the current in the coil

electromagnet

known, is shown in Fig. 34. This is the rule for the direction of the flux about a conductor backward. If tKe coil is

is

grasped in the right hand, as shown in the figure, with the

28

ELECTRICAL MACHINERY

fingers pointing in the direction that the current is flowing in around the coil, the thumb will always point toward the

\ FIG. 34.

Eight-hand rule for determining polarity of electromagnet.

north pole of the electromagnet. This is a convenient rule, often used, and an easy one to remember.

CHAPTER

III

FUNDAMENTALS OF THE ELECTRIC CIRCUIT Production of Voltage. Electricity is an invisible agent that manifests itself in various ways, and as pointed out in the introductory chapter, it is in many ways similar to heat, light and sound. Although we do not understand the true

nature of electricity, many of the laws that govern its production and application are thoroughly understood, and by the application of this knowledge very efficient electrical equipment is designed and constructed. If a

copper and a zinc plate are placed in dilute sulphuric

shown

acid, as

in Fig. 35, a difference of electrical pressure

between them that is, a current of electricity will tend to flow from the copper to the zinc plate. The terminal that the electric current tends to flow from is called the will exist

;

+

the terminal positive and is indicated by a plus sign ( ) that the current tends to flow to is called the negative and is indicated by a minus sign ( In other words, the positive ).

terminal like that it is

is

A

at a higher potential than the negative. device in the figure is called a voltaic cell or, as

shown

often referred

a great

;

to,

a primary battery, of which there are

many

types. One thing to be noted in the voltaic cell is that an electric current was not caused to flow all that was created was ;

a tendency for a current to flow from the positive to the negative terminal, or from the copper to the zinc plate. This electric pressure, or difference of potential, is expressed in a unit called a "volt," after Alessandro Volta, an Italian

who discovered the voltaic cell. This cell is sometimes called a galvanic cell, after Volta 's contemporary, Alvosio Galvani. If the terminals of the cell are connected scientist,

29

ELECTRICAL MACHINERY

30

through a copper wire or other conducting material, as shown in Fig. 36, a current will flow as indicated by the arrows.

Zinc Plate.

FIG. 36

FIG. 35

FIGS. 35 and 36.

Electric Pressure

Primary battery

and Temperature.

cells.

An

analogy of this

difference of electric pressure is found in heat. If we wish heat to flow from one body into another, a difference of

FIG. 37.

Temperature analogy of battery

cell,

Fig. 35.

temperature and also a path for the heat to flow must be For example, consider the conditions in Fig. 37.

provided.

FUNDAMENTALS OF THE ELECTRIC CIRCUIT

31

Here are two metal blocks, A and B, with a heat insulator G between them such as a firebrick or asbestos block. If heat applied to the block A, as shown, increased above that of block B, or

temperature will be might be said that a heat pressure exists between blocks A and B, just as an electric pressure existed between the copper and the zinc plates in Fig. 35 but on account of the insulation between the positive and negative terminals, which in this case is air, no current can flow. Likewise, in Fig. 37, a difference of temperature or heat pressure exists between A and B, which is expressed in deg. F. or deg. C. above or below a certain point, but no heat can flow into B because of the insulation between it and A. If, now, we remove the insulating block and replace it by a metal block, as in Fig. 38, heat will be conducted from is

its

it

;

FIG. 38.

Temperature analogy of battery

cell,

Fig. 36.

A

to B, just as an electric current flowed from the copper to the zinc plate in Fig. 36, when they were connected by a metal wire. It can hardly be said that one is any more

mysterious than the other, and in fact we do not know any .more about the true nature of one than the other. According to the electron theory of conduction in metals, the same thing takes place in both cases; that is, there is a flow of electrons from the point of high potential to that of the lower pressure. A discussion of this subject is outside the purpose of this study course; suffice it to note the close analogy between heat and electricity. Another thing is, materials that offer a high resistance to the flow of heat also offer a high resistance to the flow of electricity.

ELECTRICAL MACHINERY

32

The strength of an electric measured by the unit called an ampere, after a French scientist, Andre Marie Ampere. In the case of the heat flow from one body to the other, the quantity is measured in British thermal units (B.t.u.). When an electric curUnit of Electric Current.

current

is

rent

is passed through certain salt solutions, it will carry the metal from the terminal in the solution by which it enters and deposit it on the one by which it leaves. In Fig. 39 is

shown a

two copper plates in a copper-sulphate solution A. The current in flowing from the plate marked plus, will carry some of the copper from this plate and deposit it on the negative plate that is, the positive plate will become lighter, and the minus plate will voltaic cell C, connected to

;

Decreasing in Weight

FIG. 39.

Battery

cell

connected to electro-plating tank.

increase in weight. This is what is done to determine the standard unit of current, only silver plates are used in a nitrate of silver solution. The electric current flowing is said

ampere value when its strength is sufficient to 0.001118 deposit gram of silver per second on the negative Therefore it will be seen that there is nothing more plate. to be of one

difficult about this unit of electric current than there is about the unit of heat (B.t.u.), which is the heat necessary to raise the temperature of one pound of water one degree F. to be more accurate, from 39 to 40 deg.

Electrical

amount

Resistance.

All

substances

offer

a

certain

of resistance to the flow of heat, or in other words, tend to prevent the heat from flowing from one part of a

FUNDAMENTALS OF THE ELECTRIC CIRCUIT

33

to another. For example, in Fig. 38, when the metal block was placed between blocks A and J5, it was assumed that the heat would flow from A to B readily. In the case of Fig. 37, it was said that if a block of asbestos or firebrick

body

was placed between the metal blocks, very little heat would A to B, on account of the high heat-resisting

flow from

Similarly, all matter properties of asbestos or firebrick. offers a certain amount of resistance to the flow of elec-

Almost all good is, tends to prevent its flow. conductors of heat are good conductors of electricity. For example, copper is one of the best conductors of heat and also one of the best conductors of electricity. A very simple

tricity; that

6-Anipere

Copper

QQQQpOQQOQQoQQQQ

^^/ -Amperes

FIG. 40.

Battery

cell

connected to an iron and a copper

circuit.

experiment can be made to test the heat-conducting propercopper by taking equal lengths of copper and iron No. 10 or wire, using equal cross-sections in either case. No. 8 is a good size, and about 8 in. in length. Take one in each hand and hold each about the same distance in the flame of a Bunsen burner or blowtorch and observe which one of the ends you are holding gets hot first. It will be found that the copper will get uncomfortably hot before the iron begins to get warm, showing that the heat is conducted up the copper much faster than up the iron. The same thing is found in electricity; if equal lengths of the same size of copper and iron wire are connected to the same source of

ties of

ELECTRICAL MACHINERY

34

electricity, as in Fig. 40, it will be

times as

found that about

six

much

current will flow through the copper as the through iron, showing that copper is a much better 1 electrical conductor than iron it offers about / 6 the resistance.

No

Perfect Insulators of Electricity. On the other hand, firebrick, asbestos and other materials offer high resistance to the flow of heat. They also offer high resistance to the flow

of

electricity.

Furthermore,

insulators of heat, neither are there

there

are

no perfect

any perfect insulators

of

electricity.

The unit resistance offered by a substance to the flow of electric current is called an ohm, after Dr. Georg Simon Ohm, a German physicist. The ohm is the equivalent of the resistance offered to the flow of an electric current by a column of mercury 106.3 cm. (41.85 in.) long by 1 sq.mm. an

(0.00155 sq.in.) in cross-section at 32 deg. F. This is approximately equal to the resistance of a round copper wire 1,000

long by 0.1 in. diameter. In Figs. 37 and 38 a flame was used to create a difference of temperature between the parts A and B. This difference ft.

of temperature could have been obtained acid, although not to the same extent as

by placing A in by applying fire.

Hence we have heat produced by acid attacking the metal, just as in the voltaic cell electric pressure was produced by *he acid acting upon the zinc plate. Thus it is seen that a lifference of heat pressure, or temperature, and a difference of electric pressure can be produced in the same way. Furthermore, an electric pressure can be produced by heat. When two dissimilar metals are joined together and the joint between them is heated, a difference of electric presSuch a scheme is called a sure will exist between them. thermocouple and is illustrated in Fig. 41. Wheii used in this way, antimony and bismuth produce the highest voltage. The electric pressure produced in this manner is very small, consequently the scheme has not come into very extensive use. Nevertheless, it shows the close relation between heat

FUNDAMENTALS OF THE ELECTRIC CIRCUIT

35

and

In the introductory chapter it was pointed electricity. out that heat was produced by the friction of one body rubbing on another, and that a difference of electric pressure

was produced by a conductor moving through a magnetic field

cutting the lines of force.

FIG. 41.

Pump Analogy

Bismuth and antimony thermocouple. of Electric Circuit.

In the foregoing

we

found out that by placing a copper and a zinc plate in dilute sulphuric acid, an electric pressure was created between them, and how, if the two plates were connected together by a wire, a current of electricity would flow from the copper to the zinc. also compared the electric pressure with tem-

We

perature, and if two bodies, one of a higher temperature than the other, are in contact with each other, heat will flow

from the one at a higher temperature into the one at the lower temperature. This analogy serves a very useful purit shows that there are things that seem very simple which can be recognized only by their effects or manifestations, just as an electric pressure or current can be recognized only by its effects and manifestations. Another analogy of electric pressure and the flow of an

pose, for

ELECTRICAL MACHINERY

36

found in a pump and the flow of water this analogy lends itself most readily to illustrate many of the phenomena in an electric circuit, it will be the one most used in this and future chapters. electric current is

through a pipe.

As

In Fig. 42 is shown a centrifugal pump with the suction and discharge connected with a pipe and the system filled with water. If a valve V is placed in the pipe line and closed,

pump will not cause any water to flow. All that would do is to cause a difference of pressure between the discharge and suction of the pump, or, as it might be ex-

driving the it

FIG. 44

no. 43 FIGS. 42 to 44.

Compare

electric pressure to pressure in

a pump.

as a pressed, a tendency for the water to flow, just the same the in zinc and the plates copper pressure is created between connected not are the voltaic cell, Fig. 43. As long as plates

cannot flow. Measurement of Pressure. In heat, the temperature, or what be called the "pressure," is measured with an

externally,

an

electric current

might

instrument called a "thermometer."

pump

the pressure

is

In the case of the

measured by a pressure gage, which

FUNDAMENTALS OF THE ELECTRIC CIRCUIT shown

is

at

D

The

in Fig. 42.

pressure of the

electric

voltaic cell, or any electric device, is ment called a "voltmeter." This

37

measured by an instrument

instru-

is

shown

properly connected to measure the pressure between the positive and negative terminals of a voltaic cell in Fig. 44. The principle that the voltmeter operates on will be considered in a later chapter. To measure the pressure of an electric circuit the instruments must always be connected as in the figure namely, one terminal of the instrument connected to the positive and the other to the negative of

shown is

;

the voltaic

cell.

This

is

true in all cases;

it

does not

make

any difference whether it is an electric generator or a power If we wish to measure the pressure or lighting circuit, between the terminals, connect the voltmeters as in Fig. 44. Measurement of Current Flow. If the positive and negative terminals of the voltaic cell are connected with a wire

W,

as in Fig. 45, a current of electricity will flow, just as a

current of water would flow around in the pipe system if the valve V is open as in Fig. 47. To measure the flow of an

what may be

electric current in a circuit,

meter"

is

used.

(meaning ampere meter) and

is

is

in the circuit in Fig. 46. It is connected so that the current flowing

must also flow through the meter is connected directly terminals. If we wished to pipe line, a flowmeter would

shown

in Fig. 48.

called a "flow-

known

as an ammeter shown properly connected will be seen that the ammeter

This instrument

through the circuit

instrument, whereas the voltto the positive and negative measure the water flow in the

be connected in the system, as Comparing Fig. 46 with Fig. 48, it is at

once apparent that the flowmeter

is

connected in the pipe

[line in a way similar to that in which the ammeter is connected in the electric circuit. In either case the flow is

directly through the instrument. How to Increase Voltag'e. Since we have found out how to produce an electric pressure and current, the next thing is to

know how

to increase or decrease the quantity.

pressure of a voltaic

cell is

approximately two

The

volts, irre-

ELECTRICAL MACHINERY

38

size. Therefore, if it is desired to have a greater pressure than two volts, we will have to do something else besides change the size of the cell. A change in

spective of the

voltage

is

accomplished by connecting two or more

cells in

FIG. 48

Compare liow of

FIGS. 45 to 48.

electric current

with flow of water.

the proper relation to each other. For example, to obtain four volts, two cells of two volts each will have to be con-

nected as in Fig. 49. This is known as a series connection. The positive terminal of cell A is connected to- the negative terminal of

cell

B

and- the other positive and negative terThe arrowheads indicate that the

minals brought out.

direction of the pressure in each cell is the same. It is very essential that the cells be connected as shown in the figure.

FUNDAMENTALS OF THE ELECTRIC CIRCUIT

39

If they were connected as in Fig. 50, no pressure would be obtained between the outside terminals. This is evident when

FIG. 49.

it is

Two

battery

cells

connected in series correctly.

considered that the pressure of one

is

opposing that of

the other.

Pump Analogy of Voltage Sources in Series. Coming pump analogy, suppose the pump shown in Figs.

back to our

FIG. 50.

42,

47 and 48

Two is

battery cells connected in series incorrectly.

only capable of producing 25

Ib.

pressure.

To produce 50 Ib. pressure, two pumps would have to be connected in series, as shown in Fig. 51. In this case pump

ELECTRICAL MACHINERY

40

A

would deliver the water

to

pump B

at 25

lb.,

where

B would it

again increase the pressure by 25 lb. and discharge into the system at 50 lb. This is frequently done in high-

pressure centrifugal pumps; two or more wheels are arranged under one casing in such a way that the pressure

FIG. 51.

Pump

produced by one

analogy of two battery is

added

cells

connected

to that of the other

in series.

and

is

known

as a two-stage or three-stage

pump. To obtain more than four volts, more cells will have to be connected in series. For example, in Fig. 52, four cells are connected in series and the total pressure across them is 8 volts.

Comparing Fig. 49 with Fig. 52, it will be seen that is an enlargement of the principle involved in the

the latter

A

copper plate in one cell is connected to a zinc plate in the next, until there is only a copper and a zinc terminal remaining, and these two terminals are brought out former.

to the external circuit.

How

Current

is

Increased.

Although the pressure of a

dimensions, nevertheless the by dimensions are an important factor when the amount of current that can be taken from it is considered. The number of amperes that can be taken from a voltaic cell is almost in voltaic cell is not affected

its

direct proportion to its cubic contents; that is, a cell of 10 cu.in. volume will have about twice the ampere capacity

as one of 5 cu.in. volume, with the elements properly proporIt is not general practice to make voltaic cells very

tioned.

large in volume, therefore more than one cell must be used. To increase the current capacity and have the voltage remain the same, the cells are connected in parallel or multiple.

Note that a parallel and a multiple connection are one and the same thing.

FUNDAMENTALS OF THE ELECTRIC CIRCUIT Two

voltaic cells are

in parallel, or

be seen that the positive terhas been connected to the positive of the

multiple, in Fig. 53.

minal of one

shown connected

41

cell

It will

other, likewise for the negative terminals.

This

is

a very

simple connection and easy to remember, and the rule for may be written Connect like poles to like poles and bring

it

:

and negative lead from the group. A more common arrangement of a parallel connection is shown in Fig. 54. This is the same as in Fig. 53 that is, the positive out a positive

;

ELECTRICAL MACHINERY

42

terminals are connected together, also the negative terminals, with a positive and negative lead brought out. Pump Analogy of Current Sources in Parallel. Connect-

ing two voltaic cells in parallel is similar to connecting two pumps so that they both discharge into the same system, as

The pumps down and only one

in Fig. 55.

shut

are so connected that one

may

be

operated. If the two are running, not be any greater than when one is

would would be expected that two similar pumps of the same size would deliver twice as much water at the same pressure as one. The parallel connection is also similar to a the pressure

running, but

it

._ j

F56, 53

FIG. 57 FIG. 56 FIGS. 53 to 57. Illustrations of parallel connections.

If one is being supplied from a storage tank. used, a certain amount of water at a given pressure be drawn off before the tank will have to be filled again.

pipe line that

tank

is

may On the

other hand, if two tanks are used, the same size, and connected into the pipe line, as in Fig. 56, twice as much water will be supplied to the pipe line, but the pressure will be the same as for one tank. In connecting battery cells or electric generators in parallel it is essential that like terminals be connected together.

If unlike terminals are connected, as in Fig. 57, it

between the cells and no current through the external circuit. A connection like

will cause a closed circuit will flow

FUNDAMENTALS OF THE ELECTRIC CIRCUIT

43

would quickly destroy the cells on account of the heavy current that would circulate between them. This connection In this is similar to connecting two pumps, as in Fig. 58. case pump A will tend to cause the water to circulate around through the system in the direction of the arrow a, and pump B in the direction of arrow &. It is self-evident that opposite currents cannot flow around through the system. However, the water from A can discharge into B, and pump this

B

into A, causing a circuit through the pumps, as indicated by the arrows c, similar to the electric circuit in Fig. 57. If it is desired to connect more than two cells in parallel, is followed as in Figs. 53 and 54; like connect poles to like poles and bring out two namely,

the same general rule

FIG. 58.

Pump

analogy of incorrect parallel connection.

leads of opposite polarity. Fig. 59 shows four cells conHere the four positive terminals are nected in parallel. connected together, also the four negative terminals, with a

lead coming from the positive and another from the negative The arrowheads indicate the the external circuit L. direction of the current which flows from the positive of

to

through the external circuit and back to the negeach ative of element, and if all cells are alike they would each

cell

each deliver an equal amount of current to the circuit. There Series-Parallel and Parallel-Series Connections. are conditions that require not only a greater electric pressure than can be supplied by one cell, but also a greater current.

We

have already found out that connecting

in series increases the voltage

by the

cells

number connected.

44

ELECTRICAL MACHINERY

Also by grouping

cells in parallel, the

creased according to the

number of and

increase the voltage of the group

current

be in-

grouped. To same time make

cells so

at the

may

FUNDAMENTALS OF THE ELECTRIC CIRCUIT

45

possible to increase the current, a number of cells can be connected in a combination of a series and a parallel groupThere are two ways of doing this, first, series parallel, ing. and second, parallel series. However, either connection produces the same result. In Fig. 60 are shown four cells conit

00000000000000^

FIG. 60.

Four battery

cells

connected series-parallel.

nected in series parallel. Cells A and B are connected in Then the positive and negative terseries, also C and D. minals of group CD are connected to the positive and negative terminals of group AB respectively. By comparing Fig. 60 with Fig. 53 it will be seen that the two groups in Fig. 60 are connected the Fig. 53.

In this

way

same as the individual

cells

cells are in

can be grouped to obtain any

ELECTRICAL MACHINERY

46

voltage or current capacity desired, depending upon the number of cells so grouped. The second method is to group the cells in parallel and

then connect the groups in series, as in Fig. 61, which shows Cells A and B are cells connected in parallel series.

four

FIG. ol.

Four battery

cells

connected parallel-series.

connected in parallel, also cells C and D. Then the two groups are connected in series. It will be seen that if the connection is removed, the grouping of A and J3, and C

XY

and D, will be the same between the outside leads

The voltage and 61 is the same, but across the group it is different. In Fig. 60 two cells are connected in series in each group, which gives four volts across as that in Fig. 53.

in Figs. 60

FUNDAMENTALS OF THE ELECTRIC CIRCUIT their terminals, the

sum

of the volts of each

two

47

cells con-

series. In Fig. 61 the two cells in each group are connected in parallel; consequently the volts across the

nected in

ELECTRICAL MACHINERY

48

group

that of one

is

cell,

or in this case,

two

volts.

The flow

indicated by the arrowheads. Starting from the terminal E, the circuit is through the load L, to terminal F, where the current divides, part of it flowing of the current

is

cell A and part through cell B to point X, where the current combines again and flows to Y, where it again

through

divides, part going through C and part through D, after which it combines at E, thus completing the circuit. Fig. 62 shows the same connection as in Fig. 61, with a somewhat

arrangement of cells. It will be seen that cells and B, also C and Z>, are grouped as in Fig. 54, and then the two groups are connected in series at X and Y. By different

A

tracing out the circuits as indicated by the arrowheads, they will be found to be identical to those in Fig. 61.

As

for the results obtained

from grouping

cells,

there

preference between the series-parallel and the parallel-series connection. However, in grouping lamps or is

very

little

other devices on a circuit, the series-parallel connection is The different combinagiven preference over the latter. tions in

which elements can be grouped on an

electric circuit

are almost innumerable, but all the various groupings are only a modification of series and parallel connections; con-

sequently, by getting the idea of these two connections firmly fixed in the mind, it will greatly assist in an understanding of all other combinations.

CHAPTER IV OHM'S LAW, SERIES AND PARALLEL CIRCUITS Three Electric Units. In the chapter on elements of electricity, we became acquainted with three electric "Units and the common types of electric circuits. The three electric units were the unit of pressure, "volt"; the unit of electric' '

' '

and the unit of electrical resistance, volt corresponds to a force in pounds per square inch or a difference of temperature between two bodies or different parts of the same body. It is on just what pressure is that many misleading ideas are cultivated,

current flow,

"ohm."

in

ampere

;

The

that manifestations of pressure are mistaken for the itself, and when these manifestations are under-

pressure

considered that what pressure is, is also underMechanical pressure, the manifestations of which we so familiar with, is just as mysterious as electrical

it is

stood, stood.

are all

pressure (volts). Pressure is defined as that which tends to produce motion. It may produce motion and it may not. To '

cause motion the opposing resistance, or, as to, the reaction, must be overcome.

it is

commonly

referred

A

steam boiler under pressure looks just the same as one

and unless some external manifestation is given such as the opening of the throttle valve and an engine is put in motion, or the opening of a cock from which a jet of steam or water is given out at high velocity, we would not that

is cold,

under pressure. The same A dead circuit looks the thing same as a live one. The only way that one can be told from the other is by some external manifestation. For example, if an electric lamp is connected to a circuit and it lights, we know that an electric pressure must exist between the terbe able to tell that the boiler is

is

true of an electric circuit.

49

ELECTRICAL MACHINERY

50

minals to cause a current to flow and light the lamp. Or, again, if the terminals of the circuit are touched with the hand and a shock is received, it is realized that an electric pressure must exist to cause a current to flow and give the sensation

that

we

Therefore, in electricity as in something that is only known to exist

feel. is

mechanics, pressure by its external manifestation.

The ampere, the unit flow of same as the gallons in the capacity of a pump. For example, the state-

Unit of Electric Current. electricity, is

reference to

somewhat

inexplicit, the

ment that a given pump discharges 10,000 gal. of water a tank does not convey much of an idea as to the

into real

capacity of the pump. It may have required a month, a week, a day or an hour for the pump to deliver the 10,000 gal.

Each period would represent a different capacity pump. made that the pump discharges 10,000

If the statement is gal. into the

tank in 10 min., we then know that the

pump

has a capacity of at least 1,000 gal. per min. Therefore, to get a clear conception of what the pump is capable of doing, it is necessary to know not only the gallons or cubic feet of

but also the time that was required. The case of electricity is analogous to the foregoing. For example, if the terminals of a small battery cell are connected through an ammeter, it may set up a current of 20

fluid delivered,

or 30 amperes, but this is no indication of the capacity of the cell, for it may be capable of maintaining this rate of flow

few minutes

Or, again, a generator on a shortmay discharge an instantaneous current into the circuit of many thousand amperes, but this is no indication of

for a

only.

circuit

what the generator is capable of doing. What we are most interested in, in all commercial machines, is the number of amperes that the machine will deliver continuously. Consequently, in electricity, as in the flow of a fluid, we must consider the time as well as the quantity to get a definite idea of the capacity of a device. The unit quantity of electricity is called a coulomb, afterj the French physicist Charles Augustin de Coulomb.

The

OHM'S LAW, SERIES AND PARALLEL CIRCUITS

51

coulomb may be defined as the quantity of electricity that passes any cross-section of a circuit, in one second, when the current is maintained constant at one ampere strength. Hence, a current of electricity flowing through a circuit is measured in so many coulombs per second, just as the rate at which a quantity of fluid delivered by a pump is measured

many gallons per second, or per minute. The coulomb only the quantity of electricity as the gallon or pound is the quantity of matter. In general practice the current

in so is

strength (amperes) and not the quantity (coulombs) is'what we have to deal with; consequently, the coulomb is seldom

mentioned. Relation Between Volts, Amperes and Ohms.

was

It

pointed out in Chapter III that the three electrical units, volts, amperes and ohms, have been so chosen that one volt pressure impressed upon a circuit of one ohm resistance would cause a current of one ampere to flow. The relation

between

volts,

amperes and ohms

expression, amperes

=

volts -

ohms

;

that

may

is,

be shown by the

the current flowing in

is equal to the volts impressed on the circuit divided by the resistance of that circuit. This law was first defined by Dr. Georg Simon Ohm and may be said to be the beginning of the scientific study of electricity. The three electrical units are represented by symbols; namely, volts by E, current by / and ohms by R. Therefore,

a circuit

in

I

symbols the expression for the current

E is

I=j>>

This

rr

(formula

may

be transposed so as to read

R=

y-

;

that

is,

the

resistance of a circuit is equal to the volts divided by the current. Also, the expression may be derived, RI, which

j

E

i

indicates that the voltage impressed upon a circuit is equal ito the product of the resistance of the circuit times the cur(jrent.

i

.

How

nakes 'lave

it

to

Remember Ohm's Law.

The importance of Ohm's Law remember the foregoing formulas, or deriving them when needed. Any one of the formulas

essential that the reader

some means

for

ELECTRICAL MACHINERY

52

might be memorized and transposed to give the other two, but this Probably the easiest way requires a knowledge of elementary algebra. 77T

and remember

to express this law

it is

to write

.

RI

When

the three

symbols are arranged in this form, all that is necessary to derive any one of the three formulas is to take the element that you wish to find out of the group and place it with an equality sign to the left of the remaining two. For example, to derive a formula for finding the value

E

from the top of the group and place it, with an equality sign of E, take = RI. To obtain between, to the left of the other two elements, thus:

E

an expression

for the value of /, take / out of the

an equality sign between, to the

left of

group and place

it,

the remaining two, thus: /

with

=

E R

.

Likewise for finding the value of R, take R out of the group and place it with an equality sign between in front of the remaining two symbols, thus:

R=

E In each case the only change

.

made

in the original arrange-

'

TJJ

ment

of the

tity that

was

the equality formulas.

-

symbols

HI

was

to take the

symbol representing the quan-

to be determined out of the group and place it in front of This makes it easy to recall any one of the three

sign'.

The unit of electric pressure has a great many different names, including volts, electromotive force (e.m.f.), potential, drop of potential, difference of potential, etc., but they all mean one and the same thing. Tjl

Ohm's

law, /

=

R

,

is

a very simple expression, yet when a thorough

is obtained of it in all its applications, the subject of direct Also, the knowledge will greatly quite thoroughly understood. assist in understanding alternating current.

knowledge current

is

Electric Generator. Fig. 63 shows what is usually recog-) nized as a symbol to represent a direct-current generator or motor. It might be stated here that an electric generator and dynamo is one and the same thing. Broadly speaking, a;

generator is anything that produces something. A gas pro-; ducer is a generator of gas; a steam boiler, a generator o:

steam; and a dynamo, a generator of electricity. Howeverj the electric industry seems to have usurped the word generf ator for its own particular use, and now when we speak of *j generator, it is intended to mean a dynamo unless otherwis* specified.

OHM'S LAW, SERIES AND PARALLEL CIRCUITS In Fig. 64

53

represented a direct-current generator connected to an # = 110 means that there are 110 volts at the generator terminal and would be indicated by the voltmeter shown connected is

external circuit.

between the positive and negative side of the circuit. R = 5 indicates that the resistance of the circuit is 5 ohms, and I = 22 reads that there are 22 amperes flowing in the circuit, which should be the reading of the ammeter connected in series in the line. The values on Fig. 64 have been chosen according to Ohm's law and necessarily would have to be for them to have any meaning. By applying the formulas already given, the relation between the values will be seen. For example, suppose we know the volts E and the resistance R and wish to find the current in Then,

amperes.

by using the

expression,

FIG. 63

amperes =

volts

E_

ohms'

R'

FIG. 64

Symbol for direct-current generator. Five-ohm resistance connected to 110-volt generator.

FIG. 63. FIG. 64.

we have

7

= 5

hand, volts

To

if

=22 amperes,

the resistance

by the formula,

find resistance R,

R

volts

as indicated in the figure.

On

the other

and the current 7 are known, we may

= ohmsX amperes,

ohms =

volts ,

amperes

or /?

=

or

E = 110 = I

find the

E = RI = 5X22 = 110 volts. 5 ohms.

22

Factors that Influence Flow of Current. If 100 gal. of water per minute was flowing through a given size pipe at 50 Ib. per sq. in., by increasing the pressure to 100 Ib. per sq. in. we would expect to increase the amount. This is also true in an electric circuit, as indicated in Fig. 65. Here the conditions are the same as in Fig. 64, except that the volts E E 220 have been increased from 110 to 220. Then the current 7 = = - = 44 i

amperes, as indicated. This is true in all cases. If the pressure is doubled on a given circuit, the current will be double. At 50 Ib. per sq. in., if 100 gal. of water per minute are flowing through a pipe, of 0.5 sq. in. cross-section, then through a pipe of one-half this

ELECTRICAL MACHINERY

54

cross-section, 0.25 sq. in., we would expect to get only one-half as much or 50 gal. per min. This is also true in an electric circuit, as seen from a consideration of Fig. 66, which is the same as Fig. 64, except that the

resistance of the circuit has been doubled; that is, increased from 5 to 10 ohms. The increase of resistance may have been accomplished by

using a wire of the same material and length as in Fig. 64, but of one-half the cross-section; or the size of the wire may have remained the same as in Fig. 64, but the length doubled; or, in fact, all kinds of combimade to obtain this result. In this problem the current

nations can be ET

r\

-i -j

= 11

!= = R

10

amperes, which holds true with our water analogy.

FIG. 65

FIG. 66

FIG. 65.

Five-ohm resistance connected to 220-volt generator.

FIG. 66.

Ten-ohm

What we have

resistance connected to 110-volt generator.

seen in the example is just what would be if the resistance is doubled and the

expected; namely, that

pressure maintained constant, the current will be halved, or the current is to be decreased by 2, through a given circuit, the voltage also must be decreased by 2.

if

In the foregoing example it has been seen that it is necessary to know two of the elements to find the third that is, if the resistance of a circuit is to be calculated, it is necessary that the voltage and current be known, or if the amperes flowing in a circuit are to be determined, it will be necessary ;

to

know

the volts and the resistance of the circuit.

Ohm's Law Problems.

1.

Find the value of the

electric

that must be impressed upon the terminals of a coil of 12.5 ance to cause a current of 15 amperes to flow through it.

One lem

is

of the best

to

make

ways to

pressure

ohms

resist-

get a clear conception of an electrical probknown values on it. The con-

a diagram and put the

ditions set forth in

Problem

1

are

shown

in Fig. 67.

The problem

is

to

OHM'S LAW, SERIES AND PARALLEL CIRCUITS find the value of the volts necessary to set through a cqil of 12.5 ohms resistance. By

= 187.5 2.

A

55

up a current of 15 amperes Ohm's law, = #7 = 12.5X15 '

volts.

current of 6.6 amperes flows through an arc what is the resistance of the lamp?

lamp connected

to a

110-volt circuit;

This problem

is

shown

in Fig.

#=

E = 110 ---

3.

=16.7 ohms.

6.6

7

circuit

68 and by Ohm's law,

Determine what the ammeter readings should be, connected in a having 35 ohms resistance, when the voltmeter indicated 157.5

volts, as

shown

The

in Fig. 69.

solution

= 4.5 35 "15

=12.5

by Ohm's

law,

amperes.

Amperes

Ohms

FIG. 67

1=6.6

Amperes

E-l 10 Volts

FIG. 69

FIG. 68

FIGS. 67 to 69.

4.

A

voltaic cell has

of 1.75 volts

between

Single electric circtfits.

an internal resistance of 0.25 ohm and a pressure terminals; if the two terminals of the cell are

its

connected together by a wire of negligible resistance, what

is

the value of

the current that will flow?

The

voltaic cell

shown

in Fig. 70, has

an internal resistance from the

negative terminal around through the cell to the positive terminal of If the terminals of the cell are connected together by a short 0.25 ohm. piece of copper wire, the resistance of the latter will be so low that practically the total pressure of 1.75 volts will be effective in causing the

current to flow through the resistance of the

E

1.75

cell,

and the current

will

ELECTRICAL MACHINERY

56

What word on

Constitutes Electrical Resistance. just

what constitutes an

At

this point a

electrical resistance will

In the broad sense of the word, everything is Pieces of glass, porcelain or dry wood are resistances of extremely high value. On the other hand, a short piece of copper wire of large cross-section be in place.

an

electrical resistance.

would be a resistance of extremely low

value.

Where

a

piece of glass or porcelain of given cross-section and length may have a resistance amounting to billions of ohms, the resistance of a piece of copper of the same length and crosssection would have a resistance of only one-millionth or one-

FIG. 70.

billionth part of

Battery

an ohm.

cell short-circuited.

Nevertheless they are both

resist-

ances, the only difference being in the degree that each one opposes the flow of an electric current.

The two cases are extremes and are rarely referred to in practice as being resistances. Porcelain and glass and wood are usually called insulators, although when used on an electric circuit there is always a minute current flowing through them. The resistance of a piece of copper two or three inches in length, of ordinary cross-section, is usually spoken of as a negligible resistance, and therefore can

always be neglected in practical problems. Coils wound with copper wire of small cross-section generally have considerable resistance, depending upon the size and length of the; wire. Coils and other devices that are constructed for limiting the current through a device, or for other purposes, are;

OHM'S LAW, SERIES AND PARALLEL CIRCUITS

57

usually constructed of materials that have high resistance. High-resistance coils are generally wound of german silver or other special resistance wire.

It is

worth noting in pass-

ing that resistance wires are usually made from an alloy. Resistances that are made to carry large currents that is, comparatively low resistances such as used for starting large are usually made of cast-iron grids. From this it is seen that there may be an unlimited number of elements, all constituting a resistance to a greater or lesser degree.

motors

Series Circuits.

In the problems considered so far, only

a single element has been connected to the circuit. In practice most circuits are made up of several elements connected in various ways.

One

of the simplest

Ways of connecting group them in series. An element in a circuit may be considered as made up of a large number of sections in series. This will be made clear by reference to Fig. 71, where five separate circuits are shown different devices to a circuit is to

FIG. 71.

Five circuits in parallel.

and 5 turns respectively. If the cross-section of and the length of the turns are the same in each case, the most logical conclusion would be that the coil of two turns would have double the resistance of one turn.

of

1, 2, 3, 4,

the wire

Likewise, the one having three turns will have three times the resistance of the one having one turn, and the one of In other words, if five turns five times the resistance, etc.

elements are connected in series, the total resistance of the individual resistances so connected.

is

the

sum

The of

principle illustrated in Fig, 71 is similar to increasing the length line or the length of a coil of pipe through which a fluid is flowing.

a pipe

The longer the pipe is made, the smaller the amount of fluid that will If it is assumed that each turn of flow through it at a given pressure.

ELECTRICAL MACHINERY

58 the coil

shown

in Fig. 71

the respective coils will be

Ohm's

law, /

5, 2.5, 1.66,

=

E R

1.25

,

has

1

ohm

1, 2, 3,

to each circuit

and

1

of resistance, then the resistance of ohms as shown, and by applying

4 and 5

when

E=5

volts, the current will

be

amperes respectively, as indicated.

In Fig. 72 is shown a circuit having two resistances Ri and Jf?2 connected in series. In the preceding paragraph we found that the total resistance of a circuit consisting of two or more elements connected in series is equal to the

sum

FIG. 72.

of the individual resistances so connected.

Two

Therefore, the total resistance

R

resistances in series.

of the circuit

shown

in Fig. 72

may

be found by the formula:

= I5 ohms. The

current in amperes equals the total volts divided by the total resist-

ance; that

is,

/

=

E = 127.5 = 8.5 15 R

--

amperes.

Flow of Electric Current Compared to Water Flowing Through a Pipe. Current flowing through a circuit is just the same as water flowing through a pipe, in the respect that, what flows in at one end must flow out at the other. In the

problem, Fig. 72, 8.5 amp. leaves the positive terminal of the source and flows through the ammeter D, through resistance

then resistance R 2 back through ammeter F and into the negative terminal of the generator, the same value of current jRj,

,

entering the negative terminal that leaves the positive. Therefore the ammeter in the negative side will read the same as the one in the positive. The current is not used up in the circuit, and the same number of amperes (8.5) is effective in each resistance.

OHM'S LAW, SERIES AND PARALLEL CIRCUITS

A is

59

hydraulic analogy of the conditions existing in Fig. 72 In the latter the centrifugal pump, in Fig. 73.

shown

pressure gages, flow meters and water motors connected in series, represent respectively the generator, voltmeters, ammeters and resistances connected in series in the former.

In Fig. 73, if the whole system is full of water and the pump is driven, the water will be set in motion, and consequently the water motors, which may be used to do mechanical work. All the water that flows out of the discharge will pass

through meter D, through each motor in series, around through meter F and into the intake of the pump. Consequently each one of the flowmeters will read the same. WATER MOTORS

FIG. 73.

Hydraulic analogy of Fig.

72.

If both water motors in Fig. 73 are doing equal work and the pressure at the intake of motor No. 1 is 50 Ib. per sq.in., the most logical conclusion would be that the pressure drop through each would be equal. That is, 25 Ib. pressure would be used up in motor No. 1 and 25 Ib. in motor No. 2. If a pressure gage is placed in the pipe line between the two motors, it would read the difference in pressure between the pressure at the intake and that used up in motor No. 1, or, as we assumed in this case, 25 Ib. Likewise, a pressure gage at the discharge of motor No. 2 would read the difference between the pressure indicated on the gage between the two motors and that expended in motor No. 2. We have assumed that the remaining 25 Ib. will be used up in motor No. 2, therefore the gage on the discharge of this motor will read zero. The foregoing assumption is not absolutely correct,

because there

is

a loss of pressure in the pipe line itself. This

ELECTRICAL MACHINERY

60

feature will be given further consideration in another chapter. Nevertheless, it serves to illustrate the conditions exist-

ing around the two resistances

R

and

R2

in Fig. 72.

Voltage Drop Along a Circuit is Similar to Pressure Drop in Pipe Line. In Fig. 72 there is 127.5 volts pressure impressed upon the two outside terminals of the two resistances RI and R 2 in series. If a voltmeter A is connected across the terminals of resistance R^ it will read the amount of pressure used up in causing the current to flow through this resistance. Likewise, if a voltmeter B is connected to the terminals of resistance R 2 it will read the pressure necessary to cause the current to flow through this resistance, and the sum of the two readings should equal the total pressure ,

two resistances in series, or 127.5 volts, just as of the pressures used up in the two water motors in Fig. 73 is equal to that of the system, or that applied to the intake of motor No. 1. If both resistances were equal in across the

the

sum

A

B

and would read the same, but since Fig. 72, voltmeters are not the voltmeter they equal, readings will not be the same, the most logical conclusion being that the greatest pressure would be used in causing the current to flow through the highest resistance. A similar condition would exist in Fig. 73 if one water motor was doing more work than the other; we would expect that the greatest pressure drop would occur through the motor doing the most work.

The Value

of the Volts Across

Each Section

of Resistance in

Any

be Found by Ohm's Law, E = RI. In Fig. 72 the volts = = drop across resistance Ri may be expressed as Ei=RJ 8X8.5 68 volts. Circuit

May

The pressure across the

across

R

2

is

E = RJ = 7X8.5 = 59.5 volts.

The

z

two resistances

in series is equal to the

sum

total pressure

of the volts across

each individual resistance, or E=Ei+E2 = 68 +59. 5 = 127. 5, which checks up with the total voltage on the circuit. Right here we come in contact with a very important point in the Suppose that Ei, the reading of voltmeter A, application of Ohm's law.

and

7,

the

ammeter

reading, are

known, and

This can be done by applying Ohm's law, Tjl

expression becomes Ri

= -^ = /

R

it

is

E =

desired to find R\. or in this case

the

j,

00

^= 8.5

8 ohms

-

The same

is

true for Rz

;

that

OHM'S LAW, SERIES AND PARALLEL CIRCUITS

61

Ei 59.5 = 7 ohms. R = =

Carrying this out one step farther, suppose we 8.5 of the two resistances to be Ri=8 ohms and R2 = 7 ohms, know the value as in the figure, and then take voltmeter readings across each section and obtain for EI 68 volts, and for E2 59.5 volts. From either one of the AQ = = 8.5 amperes sets of values we may find the value of 7; that is,/ = 8 Ri -

is,

2

1

TJJ

or 7 =

= Rz

to it

know

= 8.5

amperes.

Hence

it is

evident that

it is

not necessary

i

the total resistance of a circuit and the total voltage applied to All that is necessary to know is the value of

to obtain the current.

one section of the resistance and the voltage drop across this section to determine the value of the current.

In applying Ohm's law formulas to finding the various values of the different sections of a circuit, care must be taken to use the values in the calculations pertaining to the section being considered.

on

In addition to finding the values indicated by the question marks what each instrument will read and also find the value

Fig. 74, state

FIG. 74.

Three resistances in

series.

of I by using the resistance and volts drop across each section, and check the voltage calculations by the sum of the volts across each section of resistance.

The

first

foregoing

In the step in the solution is to find the total resistance. that the total resistance of two or more resistances

we found

connected in series was equal to the sum of the individual resistances. Hence, in Fig. 74, fl = #i+.K2 +#3 = 5.5+9+2.5 = 17 ohms. The current T7F OQ K = 13.824 amperes. /=-=

R

17

ELECTRICAL MACHINERY

62

^=#,7 = 5.5X13.824=

76.03 volts.

E = RJ = 9 X 13.824 = 124.42 volts. E = RsI = 2.5X 13.824 = 34.56 volts. 2 3

The value section

of 7

7

is

=

by using the

Ei

=

=

E

2

amperes

1

;

= -^ =

meter

A

= 13.83

9

xt2

= Ra

= 13.824 2.5

amperes.

E = Ei-\-E +E

of the volts drop across the three sections is 124.42+34.56 = 235.01 volts, against 235 volts

will

ammeters

124.42

C

The sum

= 76.03 +

-

dropped and the resistance of each

- = 13.824

5.5

Ri Tfl

amperes; 7

volts

76.03

z

3

Volt-

given.

read 76.03; B, 124.42; C, 34.56; and D, 235 volts. Both will read the total current, or 13.824 amperes. In

F and G

practice the instrument could not be read much closer than the even values given, or the next higher. The most common method of grouping different Parallel Circuits.

devices on an electric circuit

FIG. 75. is

shown a simple

is

to connect

In Fig. 75

in parallel.

Single electric circuit.

circuit consisting of

a source of electromotive

them

force.

a single element R\ connected to of the current for the

The strength

-t

J7t

value of volts and resistance given

is

by Ohm's

law, 7i

=

Qf\

= Ri

= 12 15

amperes, as indicated. If a second resistance, R2 of 15 ohms is connected across the circuit, as in Fig. 76, there will also be 12 amperes ,

1

.

FIG. 76.

Two

= 24

/^rsx Am meter

equal resistances in parallel.

passing through this path, making a total of 24 amperes flowing in the main circuit. That is, 24 amperes will flow from the positive terminal of

the

generator through

the

ammeter

to

R

2

.

12

amperes flowing

OHM'S LAW, SERIES AND PARALLEL CIRCUITS

63

through R2 and the other 12 through Ri around to the negative terminal R2 where the two currents join and flow back into the negative terminal. Hydraulic Analogy of Parallel Circuit. Fig. 77 shows a hydraulic analogy of Fig. 76, where the intake and discharge of a pump are connected to two large pipes, which are connected together by several small If the system is full of water and only valve No. 1 is open, a pipes. of

,

current of water will flow depending upon the size of this pipe and the Open valve No. 2, and if this pipe is the same size as No. 1, double the amount of water will flow if the pressure is maintained constant. pressure.

FIG. 77.

Water analogy of

five circuits in parallel.

That is, a giyen quantity of water flows down pipe A, one-half of it passing through connection No. 1 and the other half through connection No. 2, where the two currents join and flow back through B to the pump; this is identical to the flow of the current in Fig. 76. The Total Resistance of Any Circuit is Equal to the Total Voltage Divided by the Total Current. In Fig. 76 the total resistance is

R=

=

= 7.5 ohms.

that of Ri or R*.

It will

be seen that this value,

In other words,

R = either

Ri or

R

2

7.5,

is

one-half

divided by

2,

the

This is true in all of resistances in parallel, in this case two. The joint resistcases where equal resistances are connected in parallel. ance of the group is equal to the resistance of one of the individual

number

by the number connected in parallel. To further illlustrate this problem, Fig. 78 is given; 20 ohms each are connected across a 120-volt circuit.

circuits divided

of

FIG. 78.

sum

By Ohm's

law,

Five lamps in parallel.

= 6 amperes. Then the total current will = 20 Ri The joint resistance of that in each circuit, or 30 amperes.

each lamp will take / equal the

here five lamps

=

ELECTRICAL MACHINERY

64

equal to the total volts divided by the total current. In this problem E 120 =4 ohms. The joint resistance is also the joint resistance R = = oU is

1.

equal to the resistance of one circuit divided by the 20 = 4 ohms. parallel, or 5

number

of circuits in

Referring again to Fig. 77, if by opening valve No. 1 six gallons per second flows around through the system, when all valves are open, if all pipes are the same size, five times six, or 30 gallons per second would flow through the system. In this analogy the pressure has been assumed to remain constant circuit

;

therefore, if five times as much water flows in the latter conditions as under the former, the

under the

resistance of the path

must have been reduced by

five.

But

the resistance offered to the flow of each individual path in Fig. 77 has not been changed. Likewise in Fig. 78 the resist-

ance of each individual circuit has not been changed; the combination of the five equal resistances in parallel only produced an equivalent resistance, equal to that of one divided by the number in parallel, or in this case one-fifth that of one individual circuit.

Joint Resistance of

Two

or More Unequal Resistances in

From

the foregoing it is evident that the problem of calculating the joint resistance of a circuit made up of a number of equal resistances connected in parallel is a

Parallel.

simple one. However, when the resistances of each individual circuit are not equal, a different method must be employed. One way of obtaining the joint resistance of a

composed of two or more unequal resistances in parallel would be to assume a given voltage applied to the circuit and then apply Ohm's law; finding the value of the current that would flow in each circuit for this voltage, and then finding the joint resistance by dividing the voltage by circuit

the total current. In Fig. 79 are shown two resistances, Ri and #2, of 20 respectively, connected in parallel across a 150-volt circuit. passing through Ri

is

h=

E = 150 = 7.5

amperes, and in

R

2

and 25 ohms The current

is

/a

=

E = 150

AND PARALLEL CIRCUITS

OHM'S, LAW, SERIES

=6

The

amperes.

and the

/i+/2 = 7.5+6 = 13.5 amperes,

current equals

total

total resistance R, equals the total volts

rent /; that

is,

R

E = =

150

1

lo 5

65

E

divided by the cur-

= 11.11 ohms.

.

Resistance and Conductance.

In reference to the flow of electricity be considered as possessing two properties first, the property which opposes the flow of an electric current (resistance), substances

all

may

and, second, the property that allows the flow of, or conducts, an electric In any material, if the resistance is high current, called conductance. conductance must be low, or conversely, if the resistance is low the con-

The

ductance must be high.

relation of the resistance to the conductance I- 13.5

o

Two

FIG. 79. of a circuit

expressed as follows:

is

the conductance

=0.05 mhos;

20

= 0.04 mhos.

= R%

is

by

R

,

and

is

usually expressed,

equal to the reciprocal of the resistance. Then, in the conductance of this circuit will equal

25

R = 25, 2

then the conductance of this circuit

The name mho, ohm

proposed for the unit of conductance. The total conductance of a circuit

1111 =

ductances connected in parallel.

ductance =

20

RZ

25

is

Then

=-5+4 =

9

100

100

1

1

RI is

represents the resistance of a

where #i=20,

79,

= /tli

R

If

the conductance will be represented

circuit,

Fig.

unequal resistances in parallel,

will

equal

spelled backward, has been

equal to the sum of the conthe problem, Fig. 79, con-

in

= 0.09 mhos.

Since conductance

the inverse of resistance, resistance must be the inverse of conductance,

that

is,

resistance

resistance,

=

.

conductance

Therefore in this problem the joint

=

R= total

conductance

9

= -^- = 11.11 ohms, which

is

the

100

same as obtained by Ohm's law. Formula for Finding Joint Resistance. The foregoing method of finding the joint resistance may be expressed by a formula. We have just seen

ELECTRICAL MACHINERY

66

that the conductance of a circuit consisting of several resistances in parallel

conductance

is,

=

(-77

Ri

The

in parallel.

R

resistance

+

>

depending upon the number

being equal to

conductance

may

be ex-

pressed as

Applying

this to

_ -_ __

our problem,

=

1

__

.

R

l

I

1

=

_ _ R-i

_t

=

1

100

---11.11 ohn*,

__

20

25

100

the same as obtained by the other two methods of working the problem. The foregoing expression will be known in the following as the jointresistance formula

In Fig. 80, in addition to finding the values indicated by the question marks, find the joint resistance of the circuit by Ohm's law; check the

FIG. 80.

results

by the

Three unequal resistances in

joint-resistance formula,

parallel.

and indicate what each ammeter

will read.

In this problem, Fig. 80,

/i

=

E = 360 = 40 E

amperes

;

360

24 amperes; ==-TT= 10 t\i

and

E = 360 = 18 "

i

+/

2

40 +24 + 18

= 82

amperes.

OHM'S LAW, SERIES AND PARALLEL CIRCUITS The

R

joint resistance

of the circuit

by Ohm's law

E 360 # =-= =4.39 By

67

is

ohms.

the joint-resistance formula, 1

1

9

the same as by the

15

1

20

Ohm's law method.

the

in the circuit

=4.39 ohms,

Ammeter B

read the current in, that is, the ammeter C will read 25 amperes, ammeter D will read 18 amperes, the cur-

the current passing through R 2 and ammeter rent flowing in R 3 ;

180

180

flowing in the branch of the circuit that current flowing through Ri, or 40 amperes;

;

=

A

will

it is

will

connected

read the total current flowing

sum

Hydraulic Analogy

of the readings of B, C, and D, or 82 amperes. of a Parallel Circuit. The conditions in the

Fig. 80, are represented

problem,

by the hydraulic analogy,

Fig. 81,

where a

pump

Pump

FIG. 81.

Water analogy of Fig.

80.

shown with its discharge and intake connected together in a pipe system. The three branch circuits, 1, 2 and 3, Fig. 81, represent respectively, The flow meters A, B, C and D in the lattei #1, fl2 and R s Fig. 80. represent respectively, the ammeters A, B, C and D in the former. If the pipe system is full of water and the pump driven, meter B will read is

,

the quantity of fluid flowing through circuit No. 1, C that flowing through No. 2, D that passing through No. 3 and A the total indicated by B, C and D, the same as ammeter A reads the total current passing through

ammeters B, C and D, Fig. 80. There can be no mistaking what an ammeter should read if the circuit is traced from the positive terminal around to the negative. In Fig. 80, starting from the positive terminal and tracing the circuit through R\, we find that the current passes through ammeters A and B, but does not

D

R

or C. 2 the Likewise, in following the circuit through current in this circuit passes through ammeters A and C, but not through B and D, that through # 3 passes through ammeters A and D, but not

go through

through

B

and

C.

In each case the current of only one circuit passes

ELECTRICAL MACHINERY

68

C and

through ammeters B,

D, where the current of

all

three circuits

D

will passes through ammeter A. Consequently ammeters B, C and read only the current of the individual circuits that they are connected in, and ammeter A reads the total of the three.

Two Methods

Working Out Joint-Resistance Problems. In the may be found by the joint-resistance

of

problem, Fig. 82, the resistance formula, 1

R=

1

1+1+1+1+1 R R* Rs

R!

R-t

3

i 6

1

,1 +1+- + 1 15

630

1?3 630

957

Five unequal resistances in parallel.

FIG. 82.

In the foregoing it was pointed out that in a simple parallel circuit, shown in the figure, the joint resistance could always be found by assuming some convenient voltage applied to the circuit and then like that

1-216.5

SI Fie. 83.

Same

as Fig. 82 connected to source of voltage.

In this case assume that the voltage equals 315,

applying Ohm's law. as indicated in Fig. 83.

Then

/!=

E = 315 = 52.5 amperes Ri

6

72 =

E = 315 =21

13 =

E = 315 =35 amperes Rz

74 =

amperes;

;

9

E = 315 =63 amperes Ri

5

and 76 =

E = 315 =45 Rf>

7

amperes.

;

;

OHM'S LAW, SERIES AND PARALLEL CIRCUITS The

current / equals the

total

sum

of

69

the currents in the individual

circuits or,

7

= / +/ +/ 3 +/4+/5 = 52.5+21+35+63 +45 = 216.5 2

1

By Ohm's law

the joint resistance

is

R= = /

amperes.

315 1.45 ohms.

This

216.5

checks up with the result obtained by the joint-resistance formula. In the problem, Fig. 84, we have two circuits in parallel, each circuit made up of two resistances in series; that is, n and r2 are connected in series forming the circuit Ri, and TZ and r4 are connected in series forming

#

the circuit

The

2.

thing to do in this problem

first

ance of each individual

is

the total resistance of a circuit composed of two or

Four unequal resistances connected

Same

FIG. 85.

connected in series

= 10

is

The

series-parallel.

as Fig. 84 connected to source of voltage.

equal to the

sum

of the individual resistances so

Hence, #i=ri+r2 = 3+5 = 8 ohms, and

ohms.

resistances

Fio. 85

FIG. 84

connected.

more

I-E7

.

FIG. 84.

to find the resist-

We found out in a previous problem that

circuit.

/22

joint resistance

= 4.44

=

R= J_

J_

Rt^Rz

ohms.

1

+ J_ 8 10

80

In this case the joint resistance may also be found by Ohm's law, as in the other problems. Assume that E = 120, as shown in Fig. 85. Then /i

=

E = 120 = 15 Ri

The is

8

total current 7

/ Volts

27

.

72

=

E = 120 = 12 amperes. Rz

= /i4-/2 = 15+12 = 27

= 4.44

R= =

and

amperes,

10

amperes, and the joint resistance

ohms, the same as obtained in the foregoing.

Drop Across a

received from a reader

The following is a question that was pertinent to these studies: What would

Circuit.

and

is

be the voltage after passing through a bank of lamps connected in series? The lamps take 0.5 ampere on a 120-volt circuit.

ELECTRICAL MACHINERY

70

In answer to this question it might be said that voltage does not pass through lamps or anything else, but is used up in causing a current In Fig. 86 are shown two water motors to flow through the circuit. connected in series in a pipe system assumed to be rilled with water, with a pump to cause the fluid to flow. If the pump is driven, it will create a certain difference in the pressure between its discharge and intake, which will cause the water to circulate and drive the motors. As was pointed out in the foregoing, if both motors are the same size, half of the pressure will be used up in causing the water to flow through motor No. 1 and half used up in motor No. 2. Or, theoretically, if we had 50 Ib. pressure at the intake of motor No. 1, the pressure would be only 25 Ib. at the intake of motor No. 2 and at its discharge, as shown. Referring to Fig. 87 and assuming that we have two such lamps as mentioned in the question connected in series across a 120-volt circuit, these lamps take 0.5 ampere on a 120-volt circuit, therefore the resistance 1 90 = = 240 ohms, as indicated on the of each lamp, by Ohm's law, is r = / 0.5 The total resistance of the circuit would be the sum of the figure. The current resistance of the two lamps, or 240+240 = 480 ohms. T/I

1 f)f\

77T

flowing through the two lamps connected in series

we

is

/

=

=

R

=0.25 480 burn dim

ampere. In this calculation why two lamps will when connected in series across a circuit having a voltage equal to the rated volts of one lamp. On the 120-volt circuit one lamp would take OA = 0.5 ampere, as mentioned in the question. But when the /= = r 240 two lamps are connected in series, the resistance of the circuit is double and the current is only one-half that taken by one alone, or in this case T77

find out

-I

0.25 ampere.

There is assumed to be 0.25 ampere flowing from the positive terminal, through the two lamps in series and back to the negative side of the line, just as a certain quantity of water was assumed to be flowing from the discharge of the pump in Fig. 86, through the two motors in series and back into the intake of the pump. In the lamp circuit it was not voltage that passed through the circuit any more than it was pressure, In each inch, that passed through the pipe system. condition the pressure was only a medium that caused a quantity to flow through a given path. Through the pipe system we call this quan-

pounds per square

water and through the lamp circuit electricity. volt drop across each lamp equals the resistance times the current, or E! = rj = 240X 0.25 = 60 volts, likewise #2 = rj = 240X .025 = 60 volts; that is, 60 volts of the pressure is used up in causing the current to flow through lamp No. 1 and 60 volts in lamp No. 2, the pressure being all expended by the time the current reaches the negative terminal of No. 2,

tity

The

OHM'S LAW, SERIES AND PARALLEL CIRCUITS

71

just as the pressure was all expended by the time the current of water reached the discharge of motor No. 2, Fig. 86. Hence, it is seen that pressure is something that does not flow, but is used up in causing some-

thing else to be transmitted from one position to another. Determining Value of Current by Volts Drop and Resistance of Section In addition to finding the values indicated by the interof a Circuit.

R

and check the answer;

is

the current

rogation marks, Fig. 88, find the joint resistance also indicate what each instrument should read.

The

first

through

and only element that we can

circuit Ri.

In the foregoing

find

we found out

that

it

/i,

flowing

was not

WATER MOTORS

FIG. 86 1-0.85

''.'Voltmeter-

FIG. 87

FIGS. 86 and 87.

Illustrate pressure drop along an electric circuit.

necessary to know the total volts and resistance of a circuit to determine All that is required is to the value of the current flowing through it. know the resistance of a section and the volts drop across this section. Referring to circuit Ri there is given the value r2 and E2 for the middle ,

E = 75 = 5 /i= 2

section

of

the

Hence,

resistance.

r2

15

amperes.

Knowing

1 = 5 amperes, n = 5 and r = 8, we may find E\ and E by Ohm's law. E = rJi = 5X 5 = 25 volts, and E = rJi = 8X 5 = 40 volts. The total volts E equals the sum of the volts drop across each section, or E = Ei+E +E = 25+75+40 = 140 volts. The total resistance R of circuit No. 1 is 2

s

1

3

l

3

2

= equal to the sum of the resistances of the three sections, or Ri r\ +rg +r = 5 15 +8 =28 ohms. The solution of this part of the problem may be checked by Ohm's law,

+

Ii=

=

=5

This checks with the value obtained by

amperes.

R

90 volts of the 140 volts total pressure is used 2 then the remainder of the 140 volts must be Hence, E t the voltage across r4 must equal the

Considering circuit

up across resistance expended across r 4.

.

rs

,

,

,

,

ELECTRICAL MACHINERY

72 difference T7F

/2

=

= r\

between

E

and

E

5

that

;

is,

ITF

prrv

= 10 5

amperes, and

r5

=

=

work by Ohm's

4

= 9 ohms.

^2=^4+^5 = 5+9 = 14 ohms. law, Rz

E 140 = = = 10

/2

volts.

The

total resistance

10

/2

of this circuit,

# = #-#5 = 140-90 = 50

O/l

We may

14 ohms, which

is

again check our

the same as obtained

by the other method.

FIG. 88.

Eesistances connected in series and the circuits in parallel.

In circuit

R h=

+/a = 5 + 10+20 = 35

Rs

7

total current

/i+/2

joint resistance

1

~

The

amperes.

The

amperes. 1

R=

By Ohm's

= -^r =20

3,

1

1

1

_

n/2

Rs

28

14

= 7

1

=

28

=4

ohms.

28

law,

E 140 R = -.= -=4 /

oo

ohms;

with the joint-resistance method. For simple parallel circuits the Ohm's law method of rinding the joint but in complex resistance is the simplest and the one generally used; circuits consisting of elements grouped in both parallel and series, it is not possible to apply Ohm's law and the conductance method has to be this checks

used.

CHAPTER V ENERGY,

WORK AND POWER

Energy the Ability for Doing Work. In considering the developed by any power-producing device, there are three elements to take into account energy, work and effort

power.

Before considering the energy transmitted by an it will be necessary to give heed to the sig-

electric current,

nificance of these three items.

Energy

is

defined as the

Air stored under pressure is an example of energy, or the ability to do work. As long as the air is confined in a tank, it cannot do any work. In order that work may be done, it may be admitted to the ability or capacity for doing work.

cylinder of a locomotive, where if the proper conditions exist, the energy stored in the air by virtue of the pressure will do the work of driving the locomotive and its load.

Again, the air may be admitted to the cylinder of a rock drill and do the work of driving the drill, or it may be used to

do the work of operating a multiplicity of different

devices.

Another example is a spring under tension ready to be released, which may, when released, do the work of moving a body from one position to another. Steam Under Pressure Possesses Energy. Steam stored under pressure in a boiler possesses energy the capacity for doing work. Unless the steam is allowed to enter the cylinder of an engine or other devices and put these devices in motion, no work is done. A horse has stored in his muscles the ability or capacity for drawing heavy loads, but as long as he is allowed to remain idle he does not do any work; nevertheless, he possesses the capacity for doing work, or energy that may be converted into work. From the fore73

ELECTRICAL MACHINERY

74

going it is seen that in order that work be done, a force must overcome a resistance that is, a body must be moved. Thus ;

work work

is is

defined as the expanding of energy, or we done when a force overcomes a resistance.

may

say

It is not enough that the throttle valve be opened and steam admitted to the cylinder of an engine that work be done, for the engine may be on dead-center or the flywheel blocked so that it cannot turn. In such a case we would have the force of the steam acting upon the piston, but no work is done because the force has not overcome the resistance offered by the mechanical connection to the piston.

All that

would occur

in the case cited

would be a

slight

increase in the space to be filled with steam. On the other hand, if conditions are such that the engine is put in motion, then work is done, or in other words, the opposition offered

by the engine and

its

load to the force of the steam has been

overcome.

Foot-Pound Unit of Work. Work is expressed as the product of a force in pounds times distance in feet. This pounds X feet == footmay also be expressed as, work in the of a force pounds X distance in feet pounds; product called the work done by a force of and foot-pounds, being one pound acting through a distance of one foot is called the unit of work, or one foot-pound. Expressing this in another way, one foot-pound is the work necessary to raise one

=

pound one

foot.

Defining Unit of Power. The simple statement that 1,000 Ib. has been raised 50 ft. that is, 50,000 ft.-lb. of work has been done does not give any idea how hard the individual

worked

from one posiFor example, suppose the weight was 10 bags of cement, of 100 Ib. each, and a workman carried them up a ladder to a floor 50 ft. from the ground. The knowledge that the task had been performed does not indicate how hard the man worked in performing the work. He may have done the job in an hour, two hours or ten hours each period would represent a different rate of doing the work. This is

or device

in transferring the 1,000 Ib.

tion to the other.

;

ENERGY, WORK AND POWER

75

where power fits in. Power is the rate of doing work, or in other words the rate of expending energy, and is equal to the

work divided by

time.

Thus; power

=

force

X

distance.

time

= foot-pound-minutes

or seconds.

The time element

in the

expressed in minutes or seconds. A foot-poundminute, the unit of power, is defined as being the work necessary to raise one pound one foot in one minute, and, the foot-pound-second, the work necessary to raise one pound

formula

is

m

one foot in one second. In the example of the

man carrying the cement, the work in either case irrespective of the same was the performed raised 50 ft., or 1,000 Ib. X 50 ft. Ib. was that is, 1,000 time; = 50,000 foot-pounds of work was performed. When we come to consider the time element, each one will represent a different rate of doing work. For example, if the man did the work in one hour (60 min.), the rate of doing the 50 distance 1,000 weight 833 foot" ,. work, power 60 time pound-minutes. If the time required is two hours (120 min.),

X

=

then the power developed minutes.

Hence

it is

=

seen that

TorT" if

is

=

= 41 ^

foot-pound-

the time element

for doing a given piece of work, the rate of doing work, is only one-half.

individual or device

X ^

is

double

power developed, the In other words, the

only working one-half as hard.

Definition of Energy, Work and Power. marize the definition of the three elements thus

We may :

Energy

sumis

the

'

capacity or ability for doing work; work is the expending of energy (this is accomplished when a force overcomes a resistance); at

and power

which energy

is

the rate of doing work, or the rate

is

expended. Horsepower Defined by James Watt. Power-developing The horsepower devices are usually rated in horsepower. was defined by James Watt, as the result of his experiments

with strong draft horses, as being the equivalent of the work necessary to raise 33,000 pounds one foot in one minute, or

ELECTRICAL MACHINERY

76

550 pounds in one second. This may be expressed as the work necessary to raise 330 Ib. 100 ft. in one minute. This unit was adopted by Watt in the rating of his steam engines and has been used since then as a unit for rating of power

=

equipment. Since the horsepower 33,000 foot-pound-minutes, the rate of doing work of any device expressed in foot-pound-minutes horsepower

=

33>OQO

When Energy

Expended Heat is Produced. Whenthat is, work done it manifests The energy expended in overcoma bearing manifests itself by heating up

is

ever energy is expended itself in the form of heat.

ing the friction of

the metal, and if the bearing is not properly lubricated to reduce the friction, the expenditure of energy necessary to overcome the friction .would be so great as to heat the metal

and destroy the bearing. If a piece of struck a blow with a hammer, it is heated here the

to the melting point

metal

is

;

energy expended in the blow is converted into heat where the blow is struck. These examples are almost unlimited. The one that is most familiar to us all is that when coal is burned in a furnace under a boiler, the energy stored in the liberated in the form of heat and passes into the converts the water into steam, which in turn transwater, mits the energy to the cylinder of the engine where it is From these exutilized to drive the engine and its load. coal

is

amples it will be seen that there is a very close connection between work and heat. In fact, work cannot be performed unless a certain amount of heat is produced. Mechanical Equivalent of Heat. After James Watt's time there flourished an English physicist, by the name of James Prescott Joule, who by experiments determined the amount of heat produced in doing a given amount of work or,

as

it

is

usually

called,

the

mechanical

equivalent

The principle of the apparatus used by Joule in his were experiments is illustrated in Fig. 89. The weights raised by turning the crank C, and then, by allowing the weights to fall, the paddle wheel P was caused to revolve in of heat.

W

ENERGY, WORK

AND POWER

77

W

the tank T, which was filled with water. The weights in pounds, minus the friction in the connected mechanism,

times the distance in feet through which they fell, would give the work done on the paddle wheel P in overcoming the

Then by taking the temperature of and after the experiment, the heat equivalent of the work expended on the paddle wheel was determined. From his experiments Joule came to the conclusion that when 772 foot-pounds of work was expended in one pound of water, its temperature would be increased one

resistance of the water.

the water before

Apparatus for determining mechanical equivalent of

FIG. 89.

heat.

degree F. Joule's apparatus was crude; consequently, as would be expected, the results were not accurate. Later experiments have shown that the correct value is 778 foot pounds. The difference between the two values is very small,

when consideration is given to the conditions under which Joule had to conduct his experiments. British Thermal Unit. The work necessary, when performed on one pound of water, to increase its temperature

one degree F. (778 foot-pounds), is called a British thermal unit (B.t.u.). Since, when 778 foot-pounds of work is performed in one pound of water, the temperature of the latter is increased one degree F., if one horsepower (33,000 foot-

pounds)

done the temperature of one pound will be

is .,

increased

33,000 I

Q
= 42.42

deg.

In other words, one horse-

is equivalent to the heat necessary to increase the temperature of one pound of water 42.42 deg.

power

ELECTRICAL MACHINERY

78

Power

of an Electric Circuit. in

an

The rate

of

which energy

(power) equal to the expended current in amperes multiplied by the volts impressed upon t he The product of volts times amperes is called circuit. watts, after James Watt, the Scottish inventor, who was responsible for much of the early development of the steam That is, electric power, watts = amperes X volts. engine. and the formula is written The symbol for watts is

is

W

electric circuit

is

W

= =

EL One

W

kilowatt,

KW,

equals 1,000 watts, therefore,

1,000-

Whenever an

electric current flows

manifests itself in the form of heat.

through a circuit, it It was by taking ad-

vantage of this fact that a connecting link was established between electrical and mechanical power. Heat Equivalent of an Electric Current. By use of an apparatus similar to that shown in Fig. 90, which shows a

FIG. 90.

Apparatus for determining heat of an

electric current.

which a coil of resistance wire is immersed, the heat equivalent of an electric current was determined. By passing an electric current through the coil it was found that when energy is expended at the rate of 746 watts, for one minute, the temperature of a given volume

vessel filled with water in

ENERGY, WORK AND POWER

79

is increased the same amount as when one mechanical horsepower of work is done upon it (33,000 ft.Ib.-min.) namely, the temperature of one pound of water is increased 42.42 deg. F. Hence we may say that 746 watts of electrical power is the equivalent of one mechanical horse-

of the water

;

power and electrical

is

called an electrical horsepower.

horsepower

Watts-

= -=;

=

W

a Since .

=-TF''

i4b

74b

w W

Therefore, -

7

El,

we

77TT

may

write, electrical horsepower

=

An electrical motor, when connected to a 220-volt circuit takes 37.5 amperes from the line. Find the power supplied to the motor in watts and electrical horsepower.

I^IGS.

91 and 92.

Elementary

circuits.

Fig. 91 shows the condition existing in the circuit in the problem; namely, a 220-volt circuit with 37.5 amperes flowing through it. The watts = = = equals volts times the current; that is, TF #7 220X37.5 8,250.

Electrical horsepower

=

W = 8,250 =11 746

746

horsepower,

and

W A

8,250

7.5-ohm resistance is connected to a 375- volt circuit. Find the in watts and electrical horsepower supplied to the resistance. The values in the problem are given in Fig. 92. Before we can find

power

ELECTRICAL MACHINERY

80

the power in watts, the current in amperes must be known. value of E and R are known, the current may be determined by Q7Pi

77?

Hence / = - =

= 50

R

watts.

amperes.

7.5

horsepower =

Electrical

The watts

law.

W = #7 = 375X50 = 18,750

W- = 18 750 = 25.3

and the

horsepower,

746

746

Since the

Ohm's

- 18.75.

W W

= EL By substituting the equivThe volts E = RI, and the watts = RIXI = RI 2 or, as it is usually RI for E in the second formula, = PR. In this problem the value of the resistance R is 7.5 written, ohms and the current 7 was calculated to be 50 amperes; then watts = I 2 R = SOX 50X7.5 = 18,750 watts. This value from the formula, checks with that obtained by the product of volts and amperes, which alent

W

,

W

shows that the expression

is

correct.

PR

The

expression (read 7 squared R) is very important, as it is generally used to express the heat losses in electrical devices, these losses

being referred to as 7 squared

R

losses (I Z

R

Another expression

losses).

ET

for the watts

is

derived as follows:

E a

the equivalent

W = ER

W = EI,

for 7 in the formula,

and 7 =

W=EI, we

R

.

have

By

substituting

E E

z

W=EX~H = H

',

2

that

is,

,

or

we may say

divided by resistance.

and the

resistance 72

Applying

= 7.5,

this to the problem, the volts 7?

E'2

then

-

that the watts equals volts times volts

W= R =

375X375

= 18,750;

the volts

W=EI

may be

when the watts and

latter the current can

for IF

= 18, 750, and

W = 18,750 =375 value of E = 7

value of 7

=W=

E

50 18 750

375

is

the

known;

E=

W for

W also 7 = E

.

obtaining

From

the

be obtained when the watts and volts are given.

-

Taking the value

transposed to read

current

is

7.5

same as obtained by the other two methods.

The formula

this

= 375,

= 50

7

= 50,

given in the problem, the

volts, as given in the

problem.

Again, the

amperes; this checks with the values given.

All these expressions are important and should be applied where possible to solving the study problems until the students have thoroughly familiarized themselves with them.

The load on a generator consists of fifty 25-watt, seventy five 60-watt incandescent lamps, five 6.5 ampere arc lamps and two motors, one

ENERGY, WORK AND POWER 5 hp. and the other 10 hp. on a 225- volt circuit.

ively

81

The motors take 18 and 30 amperes respectThe voltage of the lamp circuit is 112.5 volts.

in watts, kilowatts and electrical horsepower. If the constant for 3.5 hours, how much should a kilowatt-hour meter register if connected in the circuit during this period? The watts taken by the 25-watt lamps are equal to the number of

Find the total load

load

is

lamps times their rating, or 50X25 = 1,250; the 60-watt lamps, 75X60 =4,500 watts; the arc lamps, 5X6.5X112.5=3,656.25 watts; the 5-hp. and by the 10-hp. motor, motor, volts'X amperes = 225X18 = 4,050; 30X225 = 6,750 watts. The total watts, 1,250+4,500+3,656.25+4,050 20 206 25 = = = +6,750 20,206.25; total kilowatts, =20.2; and

W=

W

KW

total

electrical

horsepower =

W 746

1,000

= 20,206

.

25

= 27

1,000 .

1

.

The kilowatt-hours

746

equal the kilowattsXtime in hours, in this problem 20.2X3.5

= 70.7.

CHAPTER VI

COMPLEX CIRCUITS AND EFFECTS OF INTERNAL RESISTANCE Which Elements Are Grouped in Series and So far our studies have been confined to simple

Circuits in Parallel.

parallel

and

Frequently, conditions arise in

series circuits.

practice where complex circuits are involved; that is, circuits in which elements are grouped in both series and parallel.

For example, in Fig. 93, r\ and r2 are connected in parallel and the To find the combined resistance of group is connected in series with R2 such a group of elements, it is first necessary to find the joint resistance Ri of the parallel connected group. .

.

The rent

I

1

fi

f*

1

1

+ 9 7

16

16

63

# = #1+^2 = 3.94 +4.56 = 8.5 ohms. E = 115.6 /= =13.6 amperes. Knowing the current R 8.5 total resistance

we can

The

total cur-

and the

resist-

ance, By Ohm's law, the volts drop across Ri is #i=#i/ = 3.94X13.6 = 53.584 volts, and the volts drop across #2 is #2 = #2 / = 4.56X13.6 = 62.016 volts. Adding these = 53.584 +62.016 = 115.6 volts, the same results, the total volts #=#1+^2 as given in Figs. 93 and 94. The total current, 13.6 amperes, flows through R2 but divides in section The value of the Ri, part of it flowing through r\ and part through r2 current in n and r2 may be determined by Ohm's law, since we know the The current flowing in n is resistance of each branch and the volts drop. find the volts drop across each section.

,

.

ii

= Ei = 7*1

= 7.655

53.584

= 5.954

9

amperes, and the current in r2

Adding these results, the +7.655 = 13.609 amperes against 13.6. amperes.

is

i2

= Ei =

53.584 -

Tt

*

total current /=t"i+i'2

=

COMPLEX CIRCUITS

A

Internal Resistance. Fig. 93,

is

83

practical application of problem,

given in Figs. 95 and 96.

Fig. 95 shows two bat-

Each cell has a normal tery cells connected in series. internal volts and an resistance r=0.5 ohm. e=1.75 voltage, The two voltage

cells

E=

when connected

1.75

+

= 3.50

1.75

in series will give a volts.

When

normal

the cells are

connected to an external

circuit, the total pressure will not be effective in setting up a current because it will require part of the volts to cause the current to flow through the

r-

R s? '

r

I

T

Q

\

000000 -E-II5.6

FIG. 93

n=9

R 2 -4.56 1-136

E-II5B

FIG. 94

FIGS. 93 and 94.

Two

resistances connected in parallel with a third element.

internal resistance of the

that in

all

cells.

This condition

power machinery. For example,

if

and

is

in

series

similar to

in the cylinder

of a steam engine there is developed 100 hp., all the 100 hp. Part of the power will not be delivered at the flywheel.

developed in the cylinder will be used to overcome the opposition offered by the various parts of the engine itself. Likewise for a motor, if 6 hp. of electrical energy is supplied to a motor, it will not develop 6 hp. at its pulley, but only about 5 hp. This means approximately 1 hp. is required to

ELECTRICAL MACHINERY

84

overcome the friction of the moving parts of the motor and The foregoing is just the condition we have in source of electrical energy. That is, the total power any other losses.

produced in the device is not available at its terminal, as we shall see from a consideration of the latter circuit. In Fig. 96 are shown the circuit of

two

95 connected to an external

cells of Fig.

The

resistances in parallel.

joint resistance of the external

circuit is 1 1 r,

total resistance

Two

FIG. 95.

15

= 1.875

ohms.

53

ra

cells in series is R = r+r = 0.5 +0.5 = 1 R = Ri-\-Rz = 1.875 + 1 =2.875 ohms.

two

internal resistance of the

ohm, and the

=

r~i

2i

The

1

=

RI=

is

2

voltaic cells connected in series.

Tracing the circuit from the positive terminal in Fig. 96 through the resistances r\ and r z back through the cells to the positive terminal, we have a condition similar to that given in Fig. 94 allel in series

;

that

is,

two resistances connecting in parHowever, we neglected

with a third element.

the resistance of the source of electromotive force in Figs. 93 and 94 and considered the voltage constant at the terminals of the resistance,

which

is

practically so in light

and power

circuits.

How to is

Compute Value

method

is

To

of Current

When

Effect of Internal Resistance

current flowing in the circuit, the same used in Fig. 96 as in Fig. 94; namely, dividing the total volts

Considered.

find the

:

COMPLEX CIRCUITS by the volts

/=

total resistance.

of

the

battery,

E = 35 = 1.217 R 2.875

The

85

# = 3.5, the open-circuit is # = 2.875. Hence

total pressure is total resistance

and the

To

amperes.

cause the current, 7

= 1.217

amperes,

Rz = 1 ohm,

will require by Ohm's to flow through the battery resistance, law a voltage, E2 = RJ = 1 + 1.217 = 1.217 volts. This will leave available a voltage Ei at the terminals of the battery, which is equal to the total = pressure, # 3.5, minus the volts used to cause the current to flow through the battery, or #!=#-#2 =3.5 -1.217 = 2.283 volts, as indicated in the

FIG. 96.

Two

voltaic cells connected in series

ana

to

twu instances

in parallel.

applying Ohm's law to the external circuit, the current flowing T7T c\ OQO = =1.217 amperes, which is the through the external circuit is / = Ri 1.875 same as obtained by the previous method. We may now find the current

figure.

By

-

and

#x i^

in r2

.

ii=

= 2.283 =0.456 5

ampere; and 7 =ij +iz =0.456+0.761

Volts

Drop

amperes;

= 1.217

^

#1

2.283

r
3

=0.761

amperes.

in a Source of Electromotive Force.

The

volts drop in a voltaic cell or any other source of electromotive force cannot be measured directly by a voltmeter, but

must be calculated as in the foregoing. The volts drop may also be obtained by measuring the open-circuit voltage with

ELECTRICAL MACHINERY

86

a voltmeter, which in Fig. 95 would be 3.5 volts, and then measuring the voltage with the load connected as in Fig. 96.

This will give the pressure effective in causing current to flow through the load. In Fig. 96 the voltmeter should read 2.283, and the difference between these two readings gives the volts drop through the cells, or 3.5 2.283 1.217 volts. This corresponds with the calculated value previously obtained.

One feature

in these problems that stands out

more prom-

inent than any other is that it does not make any difference how the results are arrived at, when correct they will always satisfy

Ohm's

law.

A voltaic cell has a normal open-circuit voltage of 1 .8 volts. resistance

is

0.4

ohm.

Find the current that

Its internal

this cell will cause to flow

through the coil of a bell that has 2.6 ohms, also find the voltage. In the foregoing it was shown that the normal open-circuit voltage of a battery cell was not the pressure that would exist when the battery was supplying current to an external circuit, and although the electrical pressure was produced in the cell, part of this pressure is used up in causing the current to flow through the source, owing to the internal resistance. In the problem the cell in question has an internal resistance r = 0.4

ohm and

a normal voltage

FIG. 97.

E

of 1.8 volts.

shows

Fig. 97

Bell connected to voltaic

this cell con-

cell.

nected to a bell having a resistance Ri of 2.6 ohms, as called for in the problem. The total resistance of the bell and the battery cells is ET

3 ohms, the current 7 =

R

1

O

= -^- = 0.6 o

ampere, and

COMPLEX CIRCUITS

87

the voltage across the bell terminals is Ei=RJ = 2. 6X0.6 = 1.56 volts. The difference between this latter value and the open-circuit volts will give the volts e necessary to cause the current to flow through the battery; that is, e=# -#i = l. 8 -1.56 = 0.24 volt. Or, by Ohm's law, e = r/ = 0.4

X0.6 = 0.24

Here we

volt.

find that

Ohm's law may be applied equally

as well to finding the volts necessary to cause the current to flow through the internal resistance of the source of voltage as to the external circuit.

Volts

Drop Through Resistance

of

Armature.

the volts at the armature terminals

when the

the armature of a

If

generator has 0.25 ohm resistance, what voltage must a current of 35 amperes to flow through 4.75 ohms?

produce to cause Find the value of

it

current

is

watt, kilowatt, electrical horsepower supplied to the load loss in the armature.

This problem

is

flowing;

the

and the watt

worked out

The

being given in Fig. 98.

similarly to the foregoing one, the values of the circuit is the sum total resistance

R

Diagram of generator and simple

FIG. 98.

of the resistance of the armature

= 0.25+4.75 = 5 ohms. flow through the total

and the external

circuit.

circuit, or

R = r-\-Ri

The voltage necessary to cause the current to = #7 = 5X35 = 175 resistance, by Ohm's laws, is #

This total pressure is not all effective at the armature terminals, although 175 volts is what the voltmeter would read before the load is connected. A small portion of it is used to cause the current to flow through the armature. There are several ways that, volts at the armature terminals may be calculated when the current is flowing. First, by

volts.

= 166.25 volts. Second, the volts e = law, #1 =#i7 4.75X35 necessary to cause the current to flow through the armature will be equal This is found just to the resistance of the armature times the current. the same as the volts necessary to cause the current to flow through the Ohm's

In Fig. 98, e = rl = 0.25X35 = 8.75 volts, and the battery, Problem 1. difference between this and the total volts will give the volts effective at the terminals of the generator to cause the current to flow through = 175 -8.75 = 166.25 volts; this checks the external circuit, or t

# =#-e

The foregoing accounts for the voltage of a generator generally drops off a small percentage of the total volts is used applied

up with the values obtained by Ohm's law. one of the causes slightly

when load

why is

;

up to cause the current to flow through the armature. The watts W\ supplied to the load by the generator, are equal to the

ELECTRICAL MACHINERY

88

available volts EI times the current 7;

= 5,818.75;

kilowatts =

W

=

-

5,818.75

1,000

power = ture

is

W = 5,818.75 =7.8

that

is,

Wi=EJ = 166.25X35

=5.8 kilowatts and

electrical horse-

1,000

horsepower.

equal to the volts drop

e

The watts

loss

w

in the

arma-

through the armature times the current

/,

w = el = 8.75X35 =306.251 watts. Another method of obtaining the watts loss in the armature is w = I zr = 35X35X0.25 = 306.25 watts. This or

306.25 watts represents energy that has been supplied to the machine which is expended within the machine itself and appears in the form of heat.

The

total watts developed in the

armature

is

JF = #7

= 175X35

= 6,125 is

watts, and the difference between this value and the watts supplied also equal to the watts loss, or w^TF-TFj =6,125-5.818.75 =306.25

From this it is seen that 6,125 watts are developed in the armature, but only 5,818.75 are available at the load, 306.25 watts are lost in the armature, as explained in the foregoing. Effects of Increasing the Load on Batteries. Referring to the problem Fig. 97, and connecting a second bell in parallel with the first, as in Fig. 99, we will find that the volts EI will be less in the latter than in the former. The joint resistance R2 of the two bells in parallel will be one-half that watts.

/?

of one, or

R = 2

*?

A

= 22

= 1.3

ohms.

This gives a total resistance

R

for

,,

the bells and battery of 72 = fl2 +r = 1.3+0.4 = 1.7 ohms. If the cell, Fig. 99, has an open-circuit voltage, 7? = 1.8 volts as in Fig. 97, the value *

ET

of the current 7 will be 7

=

Q

= 7c

= 1.06 1.7

amperes, which will be divided

equally between the two bells in parallel, or

i

=-=

against 0.6 ampere flowing through the bells in Fig. 97. The volts EI applied to the terminals of the bells are

#!=/&/ = 1.3X1.06 = 1.378

volts,

against

1.56 volts in

=0.53 ampere,

by Ohm's Fig.

97.

law,

Also

the volts drop through the battery is e =rl =0.4X 1.06 =0.424 volt. The of these two values is # = #1+6 = 1.378+0.424 = 1.802 volts, against

sum

1.8 volts given. The difference of 0.002 volt is due to the value 1.06 amperes for the current being a little large. This is why one bell may ring satisfactorily from a battery, but when two are connected in parallel they may not ring from the same source. What we have seen in reference to Fig. 99 also accounts for the reason electric locomotives and trucks operated from storage batteries will sometimes pull more when the motors are connected in series than when they are grouped in parallel. For example, assume a condition where

the resistance of the battery is 0.5 ohm, resistance of the wires connecting the battery to the motors 0.5 ohm, open-circuit volts of the battery is 110 and the current taken by the motors when in parallel

COMPLEX CIRCUITS

89

The total resistance of the battery and connecting = ri+ri+r = 0.25 +0.25 +0.5 = 1 ohm, and the volts drop due resistance of the battery and connecting wires is E2 = RI = 1 X 100

100 amperes, Fig. 100. wires to the

is

JB

= 100 volts drop; that is, 100 volts of the total 110 are used up to cause the current to flow through the resistance of the battery and the connecting wires, which will leave only 10 volts available at the motor terminals.

The watts supplied

to the motors are

Wi= EJ = 10X100

= 1,000

watts, or 500 watts to each motor. If the motors are connected in series, as in Fig. 101, only 50 amperes will have to flow in the line to have 50 amperes flowing through the motors. With 50 amperes flowing in the line, the volts drop due to bat-

= tery and line resistance is E? RI available at the motor terminals

= 1 X 50 = 50 when they

the watts supplied to the two motors in series

FIG. 99.

Hence

watts.

the motors

it

volts.

This leaves 60 volts

are connected in series, is

and

Wi=EJ = 60X50 = 3,000

Voltaic cell connected to two bells in parallel.

will

be seen that three times the energy is supplied to are connected in series as when connected in

when they

In the first case the total watts given up by the battery is #7 = 110X100 equal to the total volts tunes the total current, or = 11,000 watts, and in the second case EI = 1 10X50 = 5,500 watts. This illustrates very clearly the necessity of keeping the resistance on parallel.

W=

W=

the source of electromotive force and the connecting wires low if a large current is to be supplied. In the first case the battery gave up 11,000 watts, but only 1,000 watts were supplied to the motors, the other 10,000 watts being lost in the connected wires and the battery, owing to their

In the second case the battery supplied a total of only 5,500 watts, but 3,000 watts, or three times that in Fig. 100, were supplied to the motors, only 2,500 watts being lost in the line and battery.

resistance.

If

the combined resistance of the line and battery had only been

ELECTRICAL MACHINERY

90

ohm, in Fig. 100, the watts supplied to the motors in parallel would be Here the con6,000, and with the motors in series, as in Fig. 101,4,250. ditions have been reversed; since the motors in parallel are supplied a 0.5

greater wattage than when connected in series, they will do a greater work. It is, therefore, evident that the internal resistance of the source of electromotive force is an important factor and must be kept low if the voltage is to remain fairly constant; also the resistance of the connecting wires must be kept low if the voltage is to remain constant at the load. These elements have a direct bearing upon the size of conductors used to transmit a given current and will be treated subsequently. 1-50

I-IOO

Fia. 100

FIG. 101

Battery connected to two motors in parallel. FIG. 101. Battery connected to two motors in series.

FIG. 100.

In Fig. 102 the battery has a normal pressure between its terminals of 6 volts and an internal resistance R\ of 0.3 ohm, which is connected to

a

The

dynamo

that develops 10 volts across

resistance of the armature

is r

its

= 0.2 ohm.

terminals on open circuit. Neglecting the resistance

of the connecting wires, find the current that will flow

through the battery

when connected

to the generator, as shown in the figure; also the volts impressed on the battery terminals. The combined resistance of the battery and armature will be = Ri+r

R

= 0.3+0.2=0.5 ohm.

The battery

is

so connected that

its

pressure will

oppose that of the generator, consequently the voltage E effective in causing a current to flow through the circuit will be the difference between that of the dynamo and the battery, or # = 10 6 = 4 volts. Note that when two electromotive forces oppose each other, only the difference between them is effective, just the same as when two mechanical forces

oppose each other.

The

rent flowing in the circuit will be / current

=

= fi

0.5

=8

amperes. Part of the 4 volts effective in causing the current to flow is used up by the resistance of the battery and part in the dynamo. The voltage required to cause the current to flow through the resistance = 2.4 volts, and this value plus the .of the battery equals 7i/=0.3X8 open-circuit voltage of the battery equals the volts, E\, impressed on its The volts drop through the armature of terminals, or #1=2.4+6=8.4.

the

dynamo

is

E = rl = 0.2X8 = 1.6 volts. 2

COMPLEX CIRCUITS

91

E

that the dynamo must develop In Fig. 103 determine the voltage 2 across the to cause 30 amperes to flow in the circuit, the volts drop s resistance R2 volts impressed on the battery terminals, watts lost in the resistance the 2 (TPi) in the armature, watts (Wz) expended

E

E

,

R

,

watts (Ws) supplied to the battery, total watts, kilowatts and electrical horsepower.

The

total resistance

R

in the circuit is the

sum

of the resistance of

the armature Ri, the external resistance R% and the battery resistance #3; = 0.15 +3.3 +0.05 = 3.5 ohms. The voltage 2 that is, # = #i 2 +#s that the dynamo will have to develop to cause a current 7 of 30 amperes

E

+#

= RI =3.5X30 = 105 to flow through the resistance R of 3.5 ohms, is r volts; 105 volts is what the dynamo would have to generate if resistance was the only opposition in the circuit, but in this case the battery has an

E

open-circuit voltage of ED = 7 5, which must also be overcome, therefore = 105+75 the dynamo will have to generate a total pressure of r -\-Eb

E= E

= 180

In the armature due to resistance there will be a voltage

volts.

103

Fia. 102

Battery connected directly to a generator. Resistance in series with a battery and generator.

FIG. 102. FIG. 103.

drop of

#a = #i/=0.15X30=4.5, = 180 7?i =727^

terminals are

and the available 4.5

= 175.5.

The

volts at the armature

voltage necessary to

overcome the ohmic resistance of the battery is E' R SI = 0.05X30 = 1.5, and the total voltage E% applied to the battery is the sum of that necessary to overcome the open-circuit volts and that to cause the current to flow through the internal resistance, or in this case, 7 3 = 7ft+7' = 75 + 1.5 = 76.5 volts. The volts drop across the resistance R* is E2 =R2 I = 3.3X30 = 99. Then a +Ez +7?3 = 4.5+99+76.5 = 180 volts, which checks

E=E

with the total previously calculated. The watts loss in the armature expended in the resistance R2 are 2

is

Wi =EJ = 4.5X30 = 135; watts

W =EJ = 99X30 = 2.970; watts supplied to the battery are W = E I = 76.5X30 = 2,295 watts, and the total watts Tf = Wi + W + TF = 135 +2,970 +2,295 = 5,400. W also equals #7 = 180 X 30 = 5,400 watts, which checks with the foregoing. The total kilowatts W 5,400 W 5,400 Z

2

3

3

and the

elecirical

7.2.

ELECTRICAL MACHINERY

92

clearly show the necescircuit that a battery is to of the the of voltage having sity be charged from approximately the same as that of the

The foregoing calculations very

the charging is to be economically done. In this problem the pressure was 99 volts too high, consequently we had to put resistance enough in the circuit to use this pres-

battery

if

sure up, or 3.3 ohms. In the resistance R 2 2,970 watts were expended which did no useful work and therefore represent a direct loss that might have been saved by having the volt,

age of the dynamo approximately that required by the battery.

CHAPTER VII ELECTRICAL INSTRUMENTS Principles on Which Electrical Instruments Are Constructed. So far attention has been called to three effects of an electric current

the magnetic effect, the heating effect

and the electrochemical effect. Electrical measuring instruments are constructed on all three of these principles. Howmost direct-current instruments, as constructed to-day employ only the magnetic effects of the current in conjunction with a permanent magnet; therefore in this discussion only electrical instruments constructed on this principle will ever,

be considered, leaving the instruments embodying the other principles for our studies of alternating current where they are mostly employed. If a coil of wire is placed so that the plane of its turns point north and south, as in Fig. 104, when a compass is placed within the coil, the needle will take a position parallel

shown. When an electric current is passed through the coil, it will cause a magnetic field to be set up about the turns of the coil, as explained in Chapter II. With the transverse axis of the coil pointing north and south, as in

to this axis, as

Fig. 104, the lines of force passing through the coil will be at right angles to the compass needle and will tend to swing

from

its position parallel to the earth magnetic result of this is shown in Fig. 105, where the needle is parallel to the lines of force, or turned 90 deg.

the latter field.

The

from the position in Fig. 104. The amount that the needle will be turned from its natural position will depend upon the strength of the current and the number of turns in the coil. This scheme is one of the simplest and oldest used for 93

ELECTRICAL MACHINERY

94

measuring

electric current, or voltage,

and

is

called a gal-

vanometer.

The next step is to reverse the order of the elements in 104 and make the magnet stationary and the coil movable. Such an arrangement is given in Fig. 106, where a horseshoe magnet is shown with a small coil suspended between its poles, the plane of the turns of the coil being Fig.

parallel to the magnetic field of the magnet.

If

an

electric

FIG. 105

^-fZ*^-

Frcs. 104 and 105.

Illustrate principle of galvanometer

caused to flow through the coil, a magnetic field up about it at right angles to the lines of force from the poles of the magnet, which Avill tend to cause the coils to revolve about the axis AB until the direction of its current

is

will also be set

magnetic

field is parallel to that of the

magnet, as in Fig.

the arrangement used in most instruments for current or voltage, and is usually known a direct measuring 107.

as

tine

This

is

movable

coil

between the poles of a permanent magnet north pole of the coil points

It will be seen that the

type. to the south pole of the

current

is

magnet. If the direction of the reversed through the coil, it would turn in the

ELECTRICAL INSTRUMENTS

95

opposite direction, consequently instruments built on this principle can be used on circuits only in which the current always flows in one direction.

~

PIG. 106

FIGS. 106 and 107.

Show

5

FIG. 107

principle of direct-current voltmeter and

.

ammeter.

Movement for Modern-Type Instrument. Fig. 108 shows a complete system of a modern-type instrument embodying To concentrate the lines of force the foregoing principle. magnet on the

coil, soft-iron polepieces I are fastened of to the pole faces the magnet M, also a soft-iron core C is supported between the polepieces, leaving only sufficient

of the

and polepieces for the coil supported on pivot bearings and

clearance between the core revolve.

The

coil

A

is

to is

opposed from turning by small spiral springs at the top and bottom of the coil; the top spring is shown at F. A pointer P made from a small aluminum tube is mounted on the coil so that when the coil moves it will be carried across the scale 8.

The

scale

8

therefore the entire system

is

supported from the magnet, be removed as a whole from

may

It will be noticed that the rightthe case for adjustment. hand terminal of the instrument is marked (positive), and the left-hand terminal -- (negative). This is standard

+

practice on all instruments of this type.

In this

way

the

ELECTRICAL MACHINERY

96

instrument

may

be used to determine the polarity of a

cir-

cuit.

The movable parts of such an instrument must be very light to reduce the friction element to a minimum, so as not to interfere with the accuracy of the indication. Consequently the coil must be wound with very fine wire, which means that it can carry only a very small current. The device shown in Fig. 108 may be used to measure either

Fia. 109

FIG. 108. FIG. 109.

Instrument for either voltmeter or ammeter. Instrument, Fig. 108, connected as a voltmeter.

voltage or current, depending upon how it is connected in the circuit. If voltage, the coil is connected in series with

a high resistance, and

if

current, in parallel with a low resist-

ance.

Instrument Connected as Voltmeter.

It

was pointed

out

in the foregoing that the movement used in a voltmeter or an ammeter is the same, the difference in the two instru-

ments being in the way they are connected to the circuit. The movement is connected in series with a high resistance when used to measure volts and in parallel with a low resistance when used to measure current. It is general practice to use about 100 ohms resistance in series with the

ELECTRICAL INSTRUMENTS

97

movable coil of the voltmeter for each volt of scale range; that is, an instrument that would indicate 100 volts on full-

=

reading will have approximately 100 X 100 10,000 ohms resistance, or an instrument that will indicate 150 volts

scale

on

=

15,000 reading will have about 100 X 150 This value will vary slightly with different Instruments that have approximately 100

full-scale

ohms

resistance.

instruments.

ohms per

volt scale reading are what may be termed the commercial type of instrument used in general general practice. However, there are types of voltmeters that have a very high resistance of 1,000,000 ohms. Such instruments are generally used for measuring high resistance such as the

insulation resistance of electrical apparatus.

In Fig. 109 is shown an instrument system similar to that in Fig. 108, connected in series with a resistance across

R

two conductors of a circuit. Assume that the resistance of the movable coil is 10 ohms and that a resistance R of 9,990 ohms is connected in series with this coil then the total

the

;

resistance of the instrument will be 10,000 ohms. If we the position on the scale and then apply a voltage

mark

E=

across the instrument I

=

the

77T

-1

=

=:

K

f\

and resistance

R

10

in series, a current

=

0.001 ampere, will be set up through 1U,UUU This current will create a turning effort against .,

coil.

the spiral springs, which, let us assume, will cause the pointer to take the position indicated at 10. By marking this position and then increasing the voltage to 20, a current Tfl

I

= -5 K

coil.

orv

=

=

0.002 ampere will be set up through the 1U,UUU This increased current through the coil will increase =

the turning effort and cause it to take a at the division marked 20 on the scale.

new

position, say

Turning Effort Proportional to the Current in the Coil. In this type of instrument the turning effort produced by the coil is directly proportional to the current flowing through it; that is, if the current in the coil is doubled the turning effort

is

doubled, consequently the distance that the

ELECTRICAL MACHINERY

98 needle will

move

across the scale will be double.

Owing

to

construction defects which are almost impossible to overcome, the foregoing is not absolutely true. If it were, slight

would be necessary

to do in dividing the scale, determine the zero and maximum points and divide the distance between these two points into the desired number of divisions. However, on the better grade of

that

all

would be

to

instruments, to eliminate any slight errors that may occur, a of cardinal points are determined by actual tests, such as the figures 0, 10, 20, 30, etc., up to 100. The scale

number

then removed, these divisions are divided, and the scale in by a draftsman. After this has been done and the scale replaced, the instrument may be used to measure volts on any direct-current circuit up to the limit of its scale. In a large percentage of voltmeters the resistance is made is

worked

form and placed inside the instrument In some portable types the resistance is made up separately and must be connected in series with the instrument

up

in a convenient

case.

when

it is

connected to the

circuit.

Not Voltage That Causes the Coil to Move, but Current. If we had followed out our calculations for the current through the instrument at different 10-volt divisions, it would have been found to be 0.001, 0.002, 0.003, etc., up to

up to the 100-volt division, not volts that actually was respectively. caused the pointer to move, but the current that was made to flow through the coil by the voltage impressed upon it and the resistance. It does not make any difference how the current may be obtained through the coil, it will move the 0.01

ampere for the

10-, 20-, 30-,

Hence, we

see

it

pointer to the position corresponding to the value of the current in the coil. For example, suppose that we remove the resistance R, as in Fig. 110, and connect the instrument 0.01. Then the current to a circuit that has a voltage E through the coil, which has a resistance of only 10 ohms, will

=

be 7

=

Tji

K

r\

=

A-|

~T?r J-U

0-001 ampere.

This value corresponds

to that obtained in Fig. 109, with 10 volts

and 10,000 ohms

ELECTRICAL INSTRUMENTS

99

resistance, consequently the pointer will throw to the same position in Fig. 110 for 0.01 volt as it did in Fig. 109 for 10 volts. In Fig. 109, when the instrument was connected to Ei

100 volts, the current through the coil was /

=0.01 ampere.

when

0.1 volt

=

=

-"

=^ rt

-i

=

r\r\

1A nr n .

1U,UUU

In Fig. 110 this value would be obtained the instrument; that is,

was impressed on

= 0.01

ampere.

In the former case the volt-

ti

meter would indicate that 100 volts was impressed across the circuit,

where

in the latter there

is

only 0.01 volt.

Instru-

I-OOOf

FIG. ill

Instrument, Fig. 108, connected as a millivoltmeter. FIG. 111. Instrument, Fig. 108, connected as an ammeter.

FIG. 110.

ments used to measure such low voltages or currents are called millivoltmeters or milliammeters, a millivolt being one thousandth of a volt, and a milliampere one thousandth

of an ampere. Many of the instruments are constructed to read in the thousandth part of a volt or ampere. Instrument Connected to Measure Current. When we

measure amperes, the instrument, as previously mentioned, is connected in parallel with a low resistance. Such a connection is

shown

in Fig. 111.

The

section of the circuit that the

ELECTRICAL MACHINERY

100 resistance

is

connected across

is

called a shunt.

The shunts

are usually constructed of a special alloy called managnin, the resistance of which is practically constant through a wide range of temperature. If the shunt 8, in Fig. Ill is

made

of such a resistance value that

when

10 amperes is flowing through the circuit 9.999 amperes will flow through the shunt and 0.001 ampere passed through the instrument's coil, then the current value in the coil will be the same as when, in Fig. 109, the instrument was connected to a 10-volt circuit, which caused the needle to be deflected to the division marked 10. Consequently, if the instrument is operating under the conditions in Fig. Ill, the pointer will move to the scale division marked 10, but instead of indicating volts the instrument will indicate amperes. The division of the current is proportional under all conditions. If the current is increased to 100 amperes, 99.99 amperes will flow through the shunt and 0.01 ampere will pass through the instrument, and the pointer will be deflected to the 100 scale division. Thus it is evident that the same movement may be used for

measuring either volts or amperes. Instrument to Measure Both Volts and Amperes. Frequently it is found advantageous, especially in portable meters, to use one instrument for both voltmeter and ammeter. When the instrument is used for a double purpose, it generally has two or more sets of values, each being marked on the scale. In the case of a voltmeter, arranged for two or more scale ranges, it is equipped with different values of resistance connected to suitable terminals. In the ammeter, it is supplied with two or more shunts of different

When

the shunts are placed inside the instrument, usually done on small-sized meters, they are connected to suitable terminals on the instrument case, and if values.

which

is

external shunts are used, as in large-capacity instruments, they are made up in suitable form to be carried around.

Where

the two instruments are combined in one, the and shunts may be located inside of small-range But in such practice the instrument must be instruments.

resistance

ELECTRICAL INSTRUMENTS

101

equipped with a key switch so as to shift from one service to Fig. 112 shows such an instrument connected to the circuit to read volts. If we assume the movable coil has the other.

20 ohms resistance, and that a 4,980-ohm resistance R is connected in series with it, then the total resistance of the instrument is 5,000 ohms. If the instrument is connected to a 50-volt

circuit

as

indicated in the figures, a current of

Fia. 112

FIG. 112. FIG. 113.

I

= -p /t

= ^-7^ OjUUU

= 0.01

which would cause

On

Fia. 113

Volt-ammeter connected to read volts. Volt-ammeter connected to read amperes.

it

ampere to

the other hand, circuit, as in Fig. 113,

move

will

flow .through the

coil,

to division 50, ;as indicated.

,

,

the instrument, is connected in tbe and the switch ^.jxreteWd agrainstrthe if

top contact a, the instrument 's coil will be connected in parallel with the shunt 8. If we assume the resistance value of

when 9.09 amperes are passing through it, 0.01 ampere will flow through the coils a total of 10 amperes in the circuit then the needle will be deflected to the same scale division with 10 amperes flowing in the circuit as when connected to read volts in Fig. 112. Howthe shunt to be such that

ever, in the latter case, the

reading will be only 10 amperes

;

ELECTRICAL MACHINERY

102

this figure is given

of the scale

is

below on the

marked

the instrument

may

for volts

scale.

Therefore

if

the top

and the bottom

for amperes, be used to indicate different values of

and amperes, depending upon how it is connected in the circuits and the position of the switch F. It is not necessary to have the two values marked on the scale only one need be used and that multiplied or divided volts

;

by 5 to obtain the other, depending upon whether the high or low values are given. Instruments that are used to read both volts and amperes are termed volt-ammeters. In some cases they are arranged so as to read three different values of amperes for example, to 1.5, to 150, liketo 15, and ;

wise three different values of volts, which may be different from the ampere values, depending upon what the instru-

ment

is to be used for. Direct-Reading Wattmeter. consideration is the wattmeter.

The next instrument There are a great

for our

many

dif-

ferent types of this instrument, but only two of them will be considered in this discussion the direct-reading wattmeter, that

is,

the instrument that indicates the product of the volts

times amperes, and the watt-hour meter. This latter device indicates the product of the volts, amperes and time in hours. This instrument

is

often incorrectly called a watt-meter.

In the wattmeter, instead of having a permanent horseshoe magnet, as in the voltmeter and ammeter, to create a flux to act upon the movable coil, two coils C are used,

widch are/coniyected in series in one side of the line, as shown in Fig. 114 V The movable coil is connected in series with a high resistance .B and across the line, the same as the coil

V

l

in Fig. 112,

when

the instrument

was connected

to read

volts. Since the coils C are connected in series in the line, they will have a current flowing through them only when a

passing through the circuit. Therefore it is only magnetic field will be produced to act upon the movable element and cause it to move. If there is no current flowing in the line, the instrument will not current

under

is

this condition that a

indicate,

no difference what value the voltage across the

coil

ELECTRICAL INSTRUMENTS

103

V may

be. For example, the instrument connected as in the 125 volts across the voltage coil, but since the line has figure is open, no current is flowing in the current coil, therefore the instrument will not indicate. This is just as it should be, since wa+ts is the product of volts times amperes. In Fig. 114 no current is flowing in the circuit, therefore the product

of volts times amperes

is

zero.

If the circuit is closed, as in Fig. 115,

and a current

of 80

amperes flows as indicated, then coils C will have a magiietic field set up in them to react upon the field of the voltage coil

FIG. 114

FIGS. 114 anO 115.

FIG. 115

Direct-reading wattmeter connected in circuit.

it to move the needle to a given position on the In this case the product of the volts and amperes is 80 10,000 watts, or 10 kilowatts. If we mark the

and cause scale.

125

X

=

position of the needle on the scale 10, as in the figure, to indicate kilowatts then by taking other readings we can

determine the different positions for various loads and work in the scale as explained for the voltmeter and ammeter in Figs. 108 to 111. The voltage of most direct-current circuits practically constant, hence the strength of coil V will be The current, however, varies as the practically constant.

is

resistance of the circuit, therefore the value of the lines of

ELECTRICAL MACHINERY

104

force will vary in coils

C and

the coil

V

will be

moved

ac-

cordingly.

An

sometimes called an electroit can be may dynamometer. used to indicate volts and amperes, and this principle is one that is used in one type of voltmeter and ammeter for use on instrument of this type

The

coils

is

be arranged so that

alternating-current circuits and will be given further consideration in our studies on this subject.

Watt-Hour Meter.

It will

be seen that an instrument of

the foregoing type reads watts direct just as the voltmeter and ammeter read volts or amperes direct respectively, and when connected to a generator or circuit, will indicate the

In the watts produced or transmitted at every instance. watt-hour meter, as previously mentioned, the instrument

The general prinregisters watt-hours or kilowatt-hours. ciple of the instrument is similar to that described for the direct-reading wattmeter, Figs. 114 and 115, except that the movable element is arranged to revolve the same as the armature of a motor. The general arrangement of one type of watt-hour meter is shown in Fig. 116. The two stationary

FIG. 116.

Diagram of watt-hour meter.

C are connected in series in the line, as in Figs. 114 and and 115, produce the magnetic field to act upon the current in the coils of the movable element V, which is constructed coils

ELECTRICAL INSTRUMENTS

105

the same as the armature of a direct-current motor, except that the winding is placed over a very light, nonmagnetic

form instead

of

an iron

The movable element

core.

is

con-

nected in series with a resistance R,_ and across the line, just as in Figs. 114 and 115. The top end of the shaft engages a registering system S, which

geared so that the pointers depending upon enlarged view of the dials is

is

will indicate watt-hours or kilowatt-hours,

the capacity of the meter.

shown

in Fig. 117.

Beginning at the

PIG. 117.

marked

10,

An

right, the first dial is

Dial for watt-hour meter.

which indicates that one revolution of

this

pointer indicates 10 kilowatt-hours, or each division indicates 1 kilowatt-hour. The next dial is marked 100, which

means that one revolution of this hand indicates 100 kilowatt-hours. The pointer on this dial moves one space for This each revolution on the hand on scale marked 10. dials. two other for the true holds system In Fig. 116,

when

current

is

flowing through the current

coils, the movable element will revolve at a speed proportional to the product of the current in coils C and the voltage impressed upon the voltage coil V. However, instead of the

hand on any particular scale .moving quickly to some division, as in Figs. 114 and 115, it will move very slowly and when the equivalent of one kilowatt for one hour has been transmitted in the circuit, the hand on the scale marked 10 will

have moved one division.

This equivalent of one kilo-

106

ELECTRICAL MACHINERY

watt for one hour may be a constant load of one kilowatt for one hour, or a constant load of 6 kilowatts for 10 minutes in fact, any combination of loads and time to produce this equivalent. In this way .the number of kilowatt-hours used in a given time is registered. A metal disk D is mounted near the bottom of the shaft

and revolves between the poles of two horseshoe magnets M, and is used to adjust the meter. When the magnets are moved out towards the edge of the disk the meter will be slowed up, where moving the magnets toward the shaft will result in an increased speed.

CHAPTER VIII

METHODS OF MEASURING RESISTANCE Voltmeter-and-Ammeter Method.

The

electrical

resist-

ance of a device or circuit can be measured in several ways. One of the simplest, if the instruments are at hand, is to connect an ammeter and voltmeter in the circuit, as shown in Fig. 118, take a reading the resistance.

and

then,

by

Ohm 's

law, calculate

AMMETER

FIG. 118.

Voltmeter and ammeter connections for measuring resistance.

For example,

if

the voltmeter reads 136.5 and the

the figure, then the resistance of the circuit

ammeter

by Ohm's law

is

R

21, as in

E 136.5 = =

= 6.5 ohms. Known-Resistance Method. A other method 'is to use a voltmeter and a resistance of known value, to determine the unknown resistance. In Fig. 119 is shown a known resistance R = 15 ohms connected in series with an unknown resistance Ri, and these two connected to a 235-volt A voltmeter is used to measure the volts drop across the terminals circuit. Assume that the values obtained are as indicated: of R and Ri, as shown. e = 75 and ei = 160. Since we know the voltage drop across the known resistance, the current

Then the unknown

By

is

determined by Ohm's law, I =

resistance R\

=

= 7 / el

107

=

75

=5

15

amperes.

=32 ohms. 5

and

substituting the values for

e

R

that

Ri

and

is,

160

32

respect-

ELECTRICAL MACHINERY

108

we have two

ively

= e\

Hence we may write the expression

equal fractions.

Written as a proportion, the formula becomes

.

Ri

e :ei

::R

:

Ri,

which indicates that the volts drop is proportional to the resistance. Therefore, to obtain any one of the values, provided the other three are is only a problem in simple proportion. Substituting the known values in the problem, Fig. 119, we have 75 15 :: 160 Ri, from which

known,

:

Ri

=

15X 160 = 75

:

32 ohms, as obtained by Ohm's law.

The foregoing is an important rule and should be remembered: When a number of resistances are connected in series across a circuit, the voltage drop across each resistance is proportional to the resistance the voltage is measured across. In Fig. 119, R : is 2.13 times as great as R; likewise, the volte^ the potential across R lt is 2.13 times as great the voltage across R. This rule holds true under all conIn using this method the voltage of the circuit must ditions.

age marked as

e,

not be so high as to destroy the resistance elements, and also of a value that will give a reasonable throw on the voltmeter. >H-

R-15

>

R,-?

AAAAAAAAAAAAAA

i A A A A A

VOLTMETERS

-E-E35

FIG. 119.

Measuring an unknown resistance with known resistance and voltmeter.

For measuring very high

Measuring High Resistance.

resistances such as the insulation resistance

of electrical

machinery or circuits, a voltmeter is used not only to measure the drop of potential, but also as the known resistance.

How

this can be done will be understood by considering 120 and 121. Figs. If

we assume that the voltmeter, Fig. 120, has a total resistance R ohms and is connected to a 100-volt circuit, the current flowing

of 10,000

through

the

instrument

is

/=

pi

.it

=

-j

rjrv

10,000

= 0.01

ampere.

With

this

METHODS OF MEASURING RESISTANCE

109

current passing through the instrument, if properly calibrated, it should indicate 100. Connecting the instrument to a circuit of only 10 volts, E 10 0.001 ampere, and the current flowing through it will be 4i 10,000 the needle will only be deflected to the 10-volt division.

/===

Instead of connecting the voltmeter to a 10-volt circuit, suppose we it in series with 90,000 ohms resistance and to a 100-volt circuit,

connect

as in Fig. 121.

The

+ 10,000 = 100,000 77f -|

J

=

is #/ = 90,000 current passing through the instrument is

total resistance of the voltmeter circuit

ohms.

The

f\(\

=

R

t

when

=0.001 ampere. This is the same value that was obtained 100,000 the instrument was connected directly to a 10-volt circuit. Conse-

FIG. 120.

Voltmeter connected to

circuit.

quently, in Fig. 121, the voltmeter will read 10 volts, the 10-volt circuit.

same

as on a

Suppose we did not know the value of Ri, but knew the resistance which in this case we have assumed to be R = 10,000 ohms.

of the voltmeter,

when the voltmeter is connected across the line as in Fig. 120, it E = IQO, and when connected as in Fig. 121, it indicates e = 10, then we have the conditions as shown in Fig. 122. Since there is only If

indicates

10-volts'

drop across the voltmeter, the remainder of the

must be expended across the unknown resistance Ri

-10 = 90

;

that

is, e\

line volts

=E

e

e\

= 100

volts.

Fig. 122 with Fig. 119, it is evident that the same conditions exist in both cases; therefore, the proportion that is true in one

Comparing

ELECTRICAL MACHINERY

110

is

also true in the other,

90 X 10,000 = 10

Hence

known

which

is:

e

:

e\

::

R

:

Ri,

from which Ri

90,000 ohms.

it is

evident that the .voltmeter can be used as the

resistance to determine the value of an

resistance as well as measure the voltage drop,

when a voltmeter

is

unknown

and

is

done

used to measure a high resistance.

FIG. 122

FIG. 121

FIGS. 121 and 122.

Show how an unknown

resistance

may be

measured with a voltmeter.

Measuring Insulation Resistance.

Frequently it is necesof electrical insulation resistance the determine sary of an for resistance the insulation apparatus example, to

armature.

This measurement

is

made

the same as explained

in Fig. 122. First connect the voltmeter across the line and determine the voltage E, then connect the voltmeter in series in one side of the line and to the commutator, with the armature shaft connected to the other side of the Assume that the readings were found as line, as shown in Fig. 123.

= 225, and when the voltmeter in the figure; namely, the line volts connected in series with the insulation of the armature, as shown,

#

is it

METHODS OF MEASURING RESISTANCE = 3.5 = 225

indicates e

volts.

ei=E

3.5

e

R = 25,000

meter to be

The

= 221.5

volts drop across the

volts.

ohms, Ri

Knowing the

=

'

=

The

armature insulation resistance of

is

the volt-

'

=1,582,143 ohms.

O.5

6

111

frequently harbored that if a circuit is insulated no current This would be true if we had a perfect Since no known perfect insulator is obtainable, it is therefore

idea

is

flows between the conductors. insulator.

impossible to insulate a circuit so that there is not at least a very small In Fig. 123 the winding current flowing from one conductor to the other. ET

is

insulated from the armature core;

nevertheless, a current 7

=

= R-\-Ri

225 J-

jOU I

j

-- = 0.00014 ampere

will

be passing through the insulation under

J.T:O

the conditions shown.

The insulation resistance of a generator or motor should be at least one million ohms.

Frequently when the ordinary

VOLTMETER R -25,000

FIG. 123.

Connections for measuring armature-insulation resistance with voltmeter.

commercial voltmeter sulation resistance

is

is connected as in Fig. 123, the in^ so high that the instrument will not

give any indication. This would indicate that the insulation was above the prescribed standard. Fig. 124 of a wiring

shows the connection for measuring the insulation resistance system. Find the insulation resistance of the conductors

from the values given. In this problem ei=E-e = 235 -2.25 = 232.75 volts and the insulation e,R 31,500X232.75 = resistance R l = =3,258,500 ohms.

--

The Wheatstone Bridge. A quently employed for measuring

piece

of

apparatus

fre-

electrical resistance is the

Wheatstone bridge. This device is generally used to measure resistances above two or three ohms. The Wheatstone '

ELECTRICAL MACHINERY

112

bridge consists of three known resistances R lt R 2 and R 3 Fig. 125, arranged so that their values can be varied by removing different plugs. galvanometer G and a battery B are con,

,

A

The bridge is closed by the unknown resistance R, which is to be determined. In measuring a resistance with the Wheatstone bridge the

nected as shown.

R 3 R 2 and R^ are so adjusted that points C and D, which the galvanometer connects to, are at the same potential. Under such a condition no current will flow through the galvanometer; consequently no deflection of the needle will be

three arms

obtained

,

,

the bridge

is

then said to be balanced.

dition like the foregoing can exist

How

where no current

a con-

will flow

VOLTMETER

R- 31.600

CONDUIT

Connections for measuring insulation resistance of conductors in a conduit*

FIG. 124.

through the galvanometer will be better understood by considering Figs. 126 and 127. In Fig. 126 the galvanometer G is connected to the terminals of resistance R 3 The current .

now

flow through the galvanometer in the direction indicated by the arrowheads. If the galvanometer is connected to as in Fig. 127, the current will the terminals of resistance 2 will

R

,

In flow through the instrument in the opposite direction. from B terminal to direction the the galvanometer moving

A

of the current has been reversed.

Instead of transferring the B to A, if the lead is terminal from galvanometer directly connected into different points in resistances R l and R a

found where no current will flow through the galvanometer this will be the point beyond which the current changes from one direction to another. At this point the voltage will be zero between the terminals of the galvanometer. location will be ;

Hydraulic Analogy of a Wheatstone Bridge.

a

fluid

A

hydraulic analogy of

Assume the two pipes 1 and 2 have flowing through them under pressure and that they are connected

this condition

is

given in Fig. 129.

METHODS OF MEASURING RESISTANCE

113

A and B by small pipes. At A the pressure is shown to be each at the points where the pipes are connected. Since the pressure is the same at these points, no fluid will flow from one pipe to the other. However, at connection B the pressure in pipe 1 is 45 Ib.,

at the points

50

Ib.

and

in

44

is 1 Ib. higher in 1 than it from 1 to 2, as indicated by the arrow. This is just the condition that we have in a Wheatstone bridge when it is balanced; the points to which the galvanometer connects are at the same pressure and no current flows through the instrument. If the bridge is not balanced, the points to which the galvanometer connects is

in pipe 2

it is

Ib.

Since the pressure

in 2, a flow will take place

ELECTRICAL MACHINERY

114

are at a different potential and a current is caused to flow through the instrument, which will be indicated by the needle being deflected.

Condition to Give Balanced Pressure. In Fig. 128 consider the bridge balanced; that is, the voltage between points C and at zero. In Fig. 129, in order that the pressure in the it will be necessary to have pipes shall be the same at point

D

B

A

the same drop in each pipe between and B; that is, there is 5 Ib. drop in No. 1, therefore there must be 5 Ib. drop in No. 2

if

no

fluid

is

to flow

from

1 to 2 at B.

In the case shown in

Fig. 129, there is 5 Ib. drop in pipe 1 and 6 consequently fluid flows from 1 to 2.

.

r*

Ib.

in pipe 2;

\50tb.

METHODS OF MEASURING RESISTANCE

115

Before we go any farther with equations (1) and (2), let us consider a simple arithmetical expression that will help to show the reason for the final process in deriving the fundamental equation for the Wheatstone bridge.

To

illustrate:

3X20-4X15,

(1)

and

3X12=4X9.

Now

if

expression

divided by

(1) is

(2),

(2)

we have

3X20 4X15

3X12~4X

9'

OA

-|

,

for the

I and

which in either

9

\2i

case equals 1.66. other to combine

p

=

In this case the 3's and 4's cancel out and leave

Likewise, two equations may be divided one by the into one. Dividing equation (1) by equation (2),

them

Wheatstone bridge, we have

/i

cancel out of each side of the equation

and the expression becomes

__ R2

Ri

R2 and R3 are known, being obtained from the bridge, Fig. 125, be found from the equation

Since R\,

R may

R= The

R,RS = 1.000X10

-^ -^T-

correctness of the formula

may

=100 ohms. be proved as follows:

Assume

= 5.5 volts. The resistance of path the voltage of the battery to be = 100 1,000 = 1,100 ohms; ECA through the bridge, Fig. 125, is

E

and the resistance of path

BDA

current flowing through path

and through path

BDA

BCA

Then the

volts drop across fl3 /i

#+^ + # +#2 = 10 + 100 = 110 3

ohms.

The

is

is

E

Rz equals

is

5.5

R equals RI = 100X0.005 = 0.5 volt,

= 10X0.05 =0.5

volt.

Hence

it

is

and across seen that the volts

ELECTRICAL MACHINERY

116

drop from point B to C and the volts drop from B to D are equal. This conforms with the conclusion previously arrived at when the bridge is

balanced,

RI = R

3 Ii.

In measuring a resistance after an adjustment of the bridge has been made, close the battery key and then the Then if care is used in closing the galgalvanometer. vanometer key, any heavy swings of the needle may be avoided if the adjustment of the bridge arms has thrown it considerably out of balance. See that the plugs are clean and do not handle them when oil or acid is on the hands. When measuring a resistance, some idea can generally be gained as to what the approximate value is; adjust the bridge to this value and close the battery and then the galvanometer key. If this is too low, try a higher value if ;

this is too great, try a value

between the two, until the

approximately balanced. It is generally not posan exact balance, but two values can be obtained, one giving a slight deflection of the needle in one direction bridge

is

sible to get

and the other causing the needle to move slightly in the opposite direction, between the two deflections the correct value lies, which can be determined very closely.

FIG. 130.

Wheatstone bridge diagram.

In Fig. 130 if the Wheatstone bridge is balanced under the conditions shown, find the value of the unknown resistance R.

The value

of

R=

flstfi H-2

= 72X1,510 =515 -I'

ohms.

CHAPTER IX CALCULATING SIZE OF CONDUCTORS Wire Gage for Round Wire.

The gage

chiefly

used in

country for measuring copper conductors is; known as the Brown & Sharpe (B. & S.) or American wire gage this

(A. W. G.) as it is sometimes called. This gage expresses the size of large conductors in their circular-mil crosssectional area, and the small-sized wires by numbers. Stand-

ard wire sizes by numbers run from No. 40 to No. 0000 any conductor larger than No. 0000 is expressed in circular-mil ;

cross-section

as

shown

in

the

accompanying

table.

A

the equivalent of the area of a circle one thousandth of an inch (0.001 in.) in diameter. On account circular-mil

is

of the small cross-section of a

number

of the wire sizes, this

small unit becomes very convenient. For example, a No. 40 wire, the smallest standard size, is only 3.145 cir.mils in cross-sectional area. Expressed in square inches the area is

0.0000077634, which at once makes apparent the advantage of using the circular mil as a unit of measurement.

Circular Mil as a Unit of Measurement. circular

By using the mil as the unit of measurement for the cross-

round wires, the area in circular mils is to the diameter in mils squared, or the diameter in equal mils is equal to the square root of the cross-sectional area in sectional area of

circular mils.

This

is

The forsquare measure is Area

evident from the following:

mula for the area of a circle in Diameter 2 X 0.7854. If the diameter

expressed in mils, the area will be in square mils. One circular mil is equal to 0.7854 part of a square mil. Hence to reduce square mils to circular mils, divide the former 117

by

is

0.7854,

from which

ELECTRICAL MACHINERY

118

Area in

circular mils

diameter 2

X

~~

= area in

0.7854

square mils

= diameter*

0.7854

Some Features

of the

Wire

Table.

The accompanying

table gives various data on conductors used for light and power installations. This table covers only wire sizes down

than

never used for light is used for the circuits running to the various equipment, Nos. 16 and 18 being used only for the wiring of lighting fixtures, to No. 18, as smaller sizes

and power

circuits.

this are

In fact no smaller than No. 14

etc. The largest-sized conductor given is 2,000,000 cir.mils in cross-sectional area. Although this is not the largest conductor made, it may be considered to be the largest

standard

larger than this being special. column headed Diameter in Mils is 500.0, and opposite this value in the left-hand column headed "Area in Circular Mils" the value 250,000 is given. This

Down

size, all sizes

' '

in the

' '

X

=

500 500, or 500 This is true of all the values in the column, "Area

latter value is obtained

by squaring

250,000. in Circular

Mils" they are equal to the values opposite, As squared, in the column headed "Diameter in Mils." 866 another example, 866 749,956, or approximately 750,000; 750,000 and 866 are found in opposite columns,

X

"Area

in Circular Mils"

=

and "Diameter

in

Mils" respec-

tively.

Wire-Number

Considering the wire-number sizes, is doubled or halved for every three numbers, depending on whether the For example, a No. 5 sizes are increasing or decreasing. wire is 33,100 cir.mils cross-sectional area taking three sizes higher, a No. 2, it will be found to have a cross-sectional area of 66,370 cir.mils 33,100 X 2 66,200 therefore a No.

it

Sizes.

will be seen that the cross-sectional area

;

;

2 wire

No.

5.

table.

=

;

approximately twice as large in cross-section as a This will be found to be the case all through the A No. 10 wire is approximately 10,000 cir.mils in

is

CALCULATING SIZE OF CONDUCTORS TABLE

6 fe

1.

119

COPPER-WIRE TABLE OF SIZES, CARRYING CAPACITIES, WEIGHT, RESISTANCE, ETC.

ELECTRICAL MACHINERY

120

cross-section area; counting

found that the area of

down

this

six sizes to No.

conductor

approximately one-quarter the

size of a

is

2583

No.

10.

16,

it

cir.mils,

is

or

By remem-

bering that the cross-sectional area of a No. 10 wire is approximately 10,000 cir.mils and that the wire sizes double

every three

sizes,

an approximate wire table may be worked

out.

Cross-Sectional and Diameter of Round Conductors. The reader should not confuse the cross-sectional area of a round conductor with the diameter. Although the crosssectional areas of round conductors double every three sizes, the diameters double only every six sizes. This is as it should be, since doubling the diameter of a circle does not increase the area by two but by four times; or in other words, the area of a circle increases as the square of the

diameter.

In the sixth column from the left-hand side of the table is given the resistance per 1000 ft. for round copper conductors at a temperature of 68 deg. F. Referring to the

No. 5 wire, per 1,000

it

ft.,

found to have a resistance of 0.313 ohm and No. 2, which is three sizes larger, has a

will be

resistance of 0.1579 ohm, or approximately one-half that of a No. 5. That is just as it should be, since the cross-sectional

area

is

doubled, which

is

the

same as connecting two equal

resistances in parallel. In this case also we must be careful to discriminate between the cross-sectional area and the

diameter of the conductor.

If the cross-sectional area is

doubled, the resistance for a given length is halved, but if the diameter is doubled, such as increasing from a No. 5 to a No. 00, the diameter of the former being 181.9 mils and that of the latter 364.8 mils or approximately double, the resist-

ance will be decreased by 4. The resistance per 1,000 ft. of No. 5 copper wire is 0.313 ohm and that of No. 00 is 0.789 or approximately one-quarter that of the former. This ;

again

is

as

it

should be

;

since the cross-section

is

increased

by four, the resistance should decrease by four, just the same as when four equal resistances are connected in parallel, the

CALCULATING SIZE OF CONDUCTORS

12.1

joint resistance of the four when connected in parallel is equal to that of one divided by four.

Wire Table Worked Out from No. to the No. 10 wire, it will be seen to

10.

Referring again have a resistance of

approximately one ohm per 1,000 ft. Remembering that the cross-sectional area doubles or halves every three sizes and that the resistance for a given length varies as the crosssectional area, an approximate table of the resistance per

The same is 1,000 ft. of round copper can be worked out. true of the weight. No. 10 bare copper weighs 31.43 Ib. per 1,000

ft.

;

No.

4,

which

is

six

numbers above No.

10,

has

approximately four times the cross-sectional area and weighs four times as much, or 127 Ibs. per 1,000 feet.

Where wires are pulled into conduits, they are usually stranded except in small sizes, and even then it is good practice to use stranded conductors as they are less liable to be broken and cause trouble after they are installed. The three right-hand columns in the table give number of strands used to make up the conductor, the diameter of each strand and the outside diameter of the conductor bare. It will be seen that the diameter of a stranded conductor is somewhat greater than that of a solid wire of the same cross-sectional area. For example, the outside diameter of a 250,000-cir.mil solid conductor is 500 mils, as given in the third column from the right, while the outside diameter of the same conductor stranded is 575 mils. This is due to the space between the strands.

A

solid conductor

can be measured with a wire gage, but

not so with a stranded conductor.

The

size of the latter

may, however, be determined by measuring the diameter of one strand in mils with a micrometer and squaring this diameter and multiplying the product by the number of For example, a No. 2 conductor is made up of 7

strands.

strands 97.5 mils diameter.

X

=

The square

of the diameter

is

9,506.25 cir.mils. The conductor is made up of 7 strands, then the total cross-sectional area is 7 9,506.25 66,544 cir.-mils, against 66,370 cir.mils given

97.5

97.5

X

=

in the table for a No. 2 conductor.

ELECTRICAL MACHINERY

122

Allowable Carrying Capacities of Copper Wire. The columns under the heading " Allowable Carrying

three

Capacities of Wires," gives the allowable load in amperes that may be placed upon a copper conductor with a rubber insulation on the wire covered with one, two or three cotton braids, also for copper wires covered with layers of varnished cotton cloth, and this insulation provided with

coverings as specified for rubber-covered wire, and also for conductors that have a weather-proof or a fire-proof insulation such as asbestos.

These carrying capacities have been

by the National Fire Protective Association, and are based upon the current that the conductor will carry established

indefinitely without increasing in temperature to where it will cause the insulation to deteriorate, and no conductors

should be loaded beyond this point. For further informaand the National Electric Code, the reader is referred to Mr. Terrell Croft's book "Wiring for Light and Power," published by the McGraw-Hill Book Co. This book deals exclusively with the application of the National tion on this table

Board of Fire Underwriters' Regulations. The carrying capacities given in the table work out

all

right for short distances, but when an attempt is made to load the conductors up to these capacities for distances over

500

found that in many cases the volts drop Therefore, it becomes necessary to base of the conductors on an allowable voltage drop and

it

ft.,

will be

becomes excessive. the size

not the allowable capacities established by the National Fire Protective Association. Of course a conductor so calculated must be at least as large for a given current as given in the table; if not, the size given in the table must be used. Allowable Voltage Drop. Just what the allowable volt-

age drop

For volts

is,

is

a question that

is

determined by conditions. more than about 3

a lighting circuit it should not be

and for motor work in general not

in excess of 5 per

cent of the line voltage, although there are cases on record where the drop in the feeders has been as high as 25 per cent.

This last value would under no condition be con-

CALCULATING SIZE OF CONDUCTORS

123

sidered good practice. In long, high-voltage transmission lines the allowable drop varies from 5 to 10 per cent, 7.5 per cent being considered a good average.

Economical Size of Conductors. The question as to just is the most economical size of conductor may appear at first thought easy of determination. This, however, has proved to be one of the problems that no general rule can be evolved for. However, one thing may be said about it As long as the returns on the power saved will pay the cost

what

:

of the additional capital invested in transmission equipment, is good engineering practice to increase the size of the

it

conductors.

many

The return on the power is affected by a great power itself, which may

factors, such as the cost of the

vary from a fraction of a cent to 8 or or the character of the installation

lOc. it

per kilowatt-hour be a new water;

may

power installation where there is an excess of power. Then no matter what may be the price per kilowatt-hour, it may be good practice to let the waterwheels do a little more work by allowing a greater drop in the line and keep the amount of capital invested in the original project down to a minibe increased as the load increases on the

mum, which may station,

by the addition of a second

line.

These and

many

other conditions enter into the most economical size of conductor in a transmission line. But for distributing circuits for lamps

and motors, the

size is generally

determined by the

voltage regulation required. Voltage Regulation of Different Circuits.

On

lighting

circuits the voltage regulation should be very close, while on motors it will depend somewhat upon the speed regulation

required, the type of motors, etc. However, as previously stated, this should not be much in excess of 5 per cent. To

determine the size of a conductor in a distributing system to transmit a given power, it is necessary to know certain factors; namely, the value of the current in amperes, the length of the circuit one way, and the allowable voltage drop.

ELECTRICAL MACHINERY

124

For example, consider a case where 175 amperes is to be transmitted ft. one way, with 3.5 per cent volts drop in the line, the voltage at the source being 236. The voltage drop in this case is 625

d

_ EX per ~

cent,

100

drop

_ 236 X 3 100

.

5

= 8.26

volts.

CALCULATING SIZE OF CONDUCTORS

125

finding a conductor that has a resistance of 0.0378 ohm per 1,000 ft. nearest resistance per 1,000 ft. that we find in the table is 0.0350.

The

This is for a 300,000-cir.mil. conductor and would be the size that must be used if the volts drop is not to exceed 8.26 when 175 amperes is passing

through the

The

circuit.

is slightly less than that calWith 175 culated, consequently the volts drop in the line will be less. amperes flowing in 1,250 ft. of 300,000-cir.mil. conductor, the volts drop

resistance of the conductor chosen

1000 ft.Xtotal length in feetXtotal cwrrenJ-MOOO, or 0.0350Xl,250X175-f-l,000 = 7.66 volts, against 8.26 calculated as the allowable. will be, resistance per

Another thing that we must keep in mind is the requirement of the National Board of Fire Underwriters' regulations, and see to it that a size of conductor is not used smaller than that prescribed in the table.

Referring to the

table for the size of conductors to carry 175 amperes, it is found to be No. 000 for rubber-covered wire, No. 00 for

for other insulations. In the case varnished-cloth, and No. of the No. 0, it is allowed to be loaded up to 200 amperes, but

No.

1,

the next size smaller, will carry only 150 amperes, will have to be used if the size of wire is

therefore No.

chosen according to the table. Taking the case of the No. 000 rubber-covered conductor, which can be installed for 175-ampere load, it has a resistance of 0.0625 ohm per 1,000 ft.,

and with 175 amperes flowing through 1,250

ft.

of this

X

X

175 -f1,250 conductor, the volts drop will be 0.0625 13.67 volts, which is considerably in excess of what 1,000

=

was decided

to be the allowable drop. Consequently the size conductor calculated will have to be used. Calculating Size of Conductors. Another method of determining the size of the conductors, in which no wire table is

required,

is

by the formula, Cir.rmls=

21 ADI

=

&d where

D

equals the distance one

already indicated.

way

in feet,

and / and Ed as

In this problem

., -. Cir.rmls =

21.4X625X175 = OQQ 283,369 5-^

ELECTRICAL MACHINERY

126

This value would be the exact

standard

conductor that will have to be used.

size of If

However, the nearest and is the

size.

300,000 cir.mils in cross-section,

size is

the resistance of a 400-ft. conductor 65,000 cir.mils. in crossis 0.45 ohm, find the resistance of 1,000 ft. of conductor of the same material 32,500 cir.mils. cross-sectional area.

sectional area

made

If the conductors were the same size in both cases, the resistance would vary directly as the length and would have been determined by

the proportion

400

:

1,000

0.45

::

:

x

or

1 000X0 45 x= -=1.125 ohm.

400

In this problem the cross-section of the conductor is decreased by 2 (from 65,000 to 32,500 cir.mils), consequently the resistance for a given length will be doubled, or the resistance in this problem is 2X1.125 = 2.25 ohms. A round conductor 750 ft. long and 350 mils diameter weighs 168 lb.; find weight of a conductor made from the same material 1,200 ft. long and 922 mils diameter. In this problem if the conductors were of the same size in both cases, the weight would vary directly as the length, and the weight of the second conductor could be determined by the proportion 750 1,200 :: :

from which

168 :x,

x=^

=268.8

lb.

The diameter

of

the

750

conductors in this problem is increased from 350 to 922 mils, hence the weight will also be increased, owing to the increased size of the conThe cross-section of a round conductor varies directly as the ductors. square of the diameter. Hence the weight of the 1,200-ft. conductor will be 350^ :922 2

The

::

268.8

x,

:

from which X

resistance of 650

0.0818 ohm.

ft.

= 1)865 3 .

o50X 350

lb

.

of copper, 83,690 cir.mils in cross-section is ft. of round copper 707 mils in

Find 'the resistance of 1,500

diameter. If

both conductors had the same cross-sectional area, the resistance of the conductor would be found by the proportion 650 1,500 :: 0.0818

1,500-ft. :

x,

:

----- =0.189

1,500X0.0818 ---

from winch x

ohm.

The diameter

of the

oou conductor

1,500-ft.

is

707 mils; then the cross-sectional area

in circular

Since mils will be diameter in mils squared, or 707X707=499,849. the cross-sectional area of the 1,500-ft. conductor has been increased

over that of the 650-ft. conductor, the resistance of the former for a given length will be decreased and will be found by the proportion QO AQQV'f) IRQ -=0.0316 ohm. 189 :x from which 499,849 83,690 499,849 :

: :

.

t

x=-

CALCULATING SIZE OF CONDUCTORS

127

it will be found that the given resistance problem was approximately that for 650 ft. of No. 1 conductor, the resistance of which is 0.1258 ohm per 1,000 ft. The conductor of which we calculated the resistance is approximately 500,000 cir.mils in

referring to the wire table

By

in the

The resistance of 1,000 ft. of 500,000-cir.mil conductor as cross-section. Then the resistance of 1,500 ft. found from the table is 0.021 ohm. would be 0.0210X1.5 = 0.0315 ohm, against 0.0316 as calculated from the resistance of the 650-ft. conductor.

conductor in circular mils required to transmit 235 allowing a drop of 4.5 per cent in the line. The voltage Also find the resistance of the line, resistance at the source is 575. per 1,000 ft. of conductor, the voltage at the load and the diameter of the

Find the amperes 875

size of

ft.,

E

conductor

The

in mils.

volts drop

Ea =

The voltage

EX - per cent drop

two conductors

*= The

.

volts.

Ea = 25.875 f ^35- =0 n E>\/

resistance per 1,000

and the

is

total length of conductor in the

Then the

= 25.875

^a =^-Ed = 575 -25.875 = 249.125,

at the load

resistance of the

575X 4 5

-100

ohm

problem 1

r\AA

f\

-

is 1

"I

ft. =

L= 875X2 = 1,750 \/

I

ft.

f\AA

- =0.0629 ohm.

L 1,750 Referring to the wire table, it is found that the nearest size conductor to one having a resistance of 0.0629 ohm per 1,000 ft. is a No. 000, which has a resistance of 0.0625 ohm per 1,000 ft. The carrying capacities given No. 000 wire is 175 amperes for rubber coyer, 270 for varnished cloth and 275 amperes for other insulation. If we assume that the conductors are to be used for inside work, and that a rubbercovered conductor is required, we cannot use the size calculated, but must use a larger wire to conform with the National Board of Fire Underin the table for a

The

a rubber-covered conductor that will 250,000 cir.mils in cross-sectional area. This rated for 240 amperes and has a resistance of 0.042 ohm per

writers' regulations.

meet these requirements conductor

is

size of

is

resistance of 1,750 ft. will be 1,750X0.042-M, 000 = 0.0735 ohm, and the volts drop when 235 amperes is flowing through the circuit is = #7 = 0.0735X235 = 17.27, against 25.874 volts that was allowed. #
1,000

ft.

The

nearly constant at the load with variations in the value of the current. The size of conductor to transmit 235 amperes 875 ft. with 25.875 volts

drop in the

line

may

also

be found by the formula,

21.4D7

, 21.4X875X235 -,170,000. 25 ^ 75

ELECTRICAL MACHINERY

128

which checks up very closely with the cross-sectional area of a No. 000 conductor as given in the table. The diameter of a round conductor in mils

equal to the square root of the cross-sectional area in circular mils, our problem, diameter in mils = VlTOjOOO = 412.

is

or, in

Derivation of Circular-Mil Formula.

The formula for

finding circular-mil area of a conductor to transmit a given current is based on the resistance of 1 circular-mil-foot of

A

circular-mil-foot is a wire 1 ft. long and 1 cir.mil copper. in cross-section, or in other words a round wire 1 ft. long by one thousandth of an inch (0.001 in.) diameter. The resist-

ance of 1

cir.mil-ft. of

copper

is

given different values for

different temperatures, varying from 10.5 to 11 ohms, 10.7 ohms being a good average value. The foregoing statement

means that

if

a round section 1

ft.

long and 0.001

in.

diameter is taken from any copper wire, it will have a resistance of 10.7 ohms at a temperature of about 70 deg. F. If we take a 1-ft. section of a conductor and consider this made up of a number of wires in parallel, each element 1 cir.mil in cross-sectional area, each one of these little wires will have a resistance of 10.7 ohms, and if the cross-sectional area of the conductor is equivalent to 100 cir.mils, the resistance of the 1-ft. section will be 10.7 -f- 100 = 0.107 ohm. Looking at this in another way, if the conductor is 100 cir.mils in cross-section, we have a condition similar to that if the wire was made up of 100 wires each 1 cir.mil in crosssection. In this case the same law will apply as when connecting equal resistances in parallel; that is, the joint resistance of the group would equal the value of one resistance divided by the number connected in parallel. In our problem we considered that the 1-ft. section of conductor was

made up

of 100 wires 0.001

in.

in diameter

and that each wire

Since there are 100 in par10 7 or 0.107 ohm. If the the joint resistance will be

had a resistance of 10.7 ohms.

_

'

allel,

cross-sectional area of the

1-ft.

}

section

was equivalent to would be

1 000 cir.mils, then the resistance of the piece 10 7-:- 1,000 0.0107 ohm. ?

=

CALCULATING SIZE OF CONDUCTORS

129

Resistance per Foot of Conductor. From the foregoing evident that the resistance of a 1-ft. section of any copper conductor is equal to 10.7 divided by the circular-mil it is

Take, for example, a No. 8 wire. Its resistper 1,000 ft. This is equivalent to 0.000641 foot. The sectional area of a No. 8 wire is

cross-section.

ance

ohm

0.641

is

ohm per

foot equals 10.7

then the resistance

cir.mils

=

per ~ 16,510 - 0.000648 ohm, against 0.00641 ohm given in the

16,510 10.7

;

-*-

foregoing. As another example, take a 300,000-cir.mil conductor, the resistance of which is 0.035 ohm per 1,000 ft., or 0.000035

ohm per foot

=

From

foot.

10.7

-f-

the method described, the resistance per 0.0000356 ohm, 10.7 -f- 300,000

=

cir.mils

=

which checks up fairly close to that given in the wire table. Since we have a method for determining the resistance per foot of any copper conductor, all that is necessary is to multiply the result thus obtained by the length of the conductor in feet to find the total resistance. If we denote the length of the conductor by L, then the resistance of the con-

ductor

is

R=

r^

~-

cir. mils'

To illustrate, the resistance of 1,750 ft. of No. 2 conductor 0.276 ohm, as calculated from the resistance per 1,000 ft. given in the wire table, and the cross-sectional area is 66,370

is

cir.mills.

the

Calculating

from the

resistance

foregoing

formula, 10.7L

10.7X1,750

aai

cir.mils

against 0.276

ohm

A000 0.282 ohm ,

66,370

as calculated

from the data given in the

wire table. Since circular mils

is

one of the elements in the formula

for finding the resistance R, the expression

read cir.mils

=

'-

for

finding

the

may

circular

be

made

mils.

to

The

1

lengths of circuits are usually expressed in the distance one way, and not in the total length of wire in them, and the value of the volts drop and current transmitted are more readily at

130

ELECTRICAL MACHINERY

-

hand than the

resistance of the circuit.

express the length

by 2D, where

L

Therefore,

we

will

of the conductor in the two-wire circuits

D equals the distance one way,

and the

resistance

ET

of the conductors

shown

as already

is

stituting these equivalents for expression becomes

Cir.mils =

to be,

L and R

R =-y

By

.

sub-

in the formula, the

10.7X2D = 21.4D7 = j= &d

J^d

T

which

is

we

the formula

the foregoing.

use to find the size of conductors in

The formula may be transposed

to read

:

ADI E ^21 cir.mils

for finding the volts drop in a given circuit, or to read

j_cir.milsXEd

21AD

for finding the current that may be transmitted through a given circuit with a given allowable voltage drop. In a circuit 750 ft. long, one way, in which 350,000-cir.mil conif the voltage at the load is 230 when 175 amperes is being transmitted, find the volts drop in the line, the volts at the source of power, the resistance of the conductor, the total watts supplied to the line, the watt loss in the line and the watts supplied to the load. 1.

ductors are used,

The

volts drop in the line, in this problem,

d

~

21.4D7

~ 21.4X750X175

cir.mils

The

volts

E

at the source of

the load and the volts drop the line

is

R=

= L

,

350,000

power equal the sum

or

of the volts

E = E a +Ed = 230 +8 = 238.

Ea

at

Resistance

O

77T

of

Ed

is

=0.045 ohm. 175

Figuring another way,

the

resistance of the line will be the resistance of 1,500 ft. of 350,000-cir.mil conductor; referring to the wire table, this size conductor will be

ohm per 1,000 ft., or 65X0.03=0.045 supplied to the system are equal per 1,500 ft. The total watts = El = 238X175 to the total volts E times the total current I, that is, found to have a resistance of 0.03

ohm

W

W

CALCULATING SIZE OF CONDUCTORS

W in the line equals the volts drop Ed times the E = 8X175 = 1,400; the watts Wa supplied to the the volts E a at the load times the current 7; hence

= 41,650; watts loss current 7, or Wi = dI load are equal to

131

t

W = EJ = 230X175 = 40,250, a

and the sum

of the watts lost

and those

W = Wi+W

supplied to the load should be equal to the total watts, or 1,400+40,250 = 41,650. This checks with the first method.

a

=

A 1,000-kw 120-volt compound-wound generator furnishes current to a lamp load at a distance of f mile and is adjusted to maintain constant voltage at the load. The voltage drop in the line at full load is chosen What size wire is required to be 10 per cent, of the normal voltage. What percentage of its full-load if 2,100 forty-watt lamps are turned on? capacity would the generator be delivering? What would be the voltage at the generator? What would be the watts loss in the line? What percentage of the load supplied would the line loss be? First, it was required to find the size of wire required to limit the line drop to 10 per cent, of the normal voltage. Since 120 volts is the normal voltage, the voltage drop in line would be 10 per cent, of 120 volts or y^X 120 = 12 volts. The resistance of the line would be the voltage drop in it divided by the current flowing through it. The current flowing would be the watts taken by the load divided by the voltage. The load consists of 2,100 forty-watt lamps and would therefore be 2,100X40 = 84,000 watts. Consequently the line current would be this number of 84L000 =700 amperes. The resistance of watts divided by 120 volts, or the line would therefore be the voltage drop divided by this current, or 12 =0.01714 ohm. There are two lines each f mile long, making a total

700

length of f mile. Since there are 5,280 ft. in a mile, the total length of wire in the line is |X 5,280 = 3,960 ft. The resistance per thousand feet would be the total resistance of the line divided, by its length in feet

-

01714

multiplied

thousand

From

by one thousand, or

=1,000 = 0.004329

ohm

per

feet.

found that a 2,000,000-cir.mil cable has a per 1,000 ft., which is greater than the resistance required. Since this is the largest cable in general use, it would be necessary to use two cables in multiple to meet the conditions. We could use one cable of 2,000,000 cir.mil and a smaller one of such size the wire table

resistance of 0.005177

it

is

ohm

as to reduce the resistance to the value desired, but the more usual way would be to use two like cables of such size as would give the required resistance when placed in multiple. resistance are placed in multiple the resistance of either one, which is the of either is twice their

combined

When two conductors of the same combined resistance is one-half the same as saying that the resistance

resistance.

In the case of our problem

ELECTRICAL MACHINERY

132

the combined resistance resistance of each of the

is

to be 0.004329

ohm

per 1,000

ft.;

hence the

two conductors would be twice that

value, or From the wire table we find 1,000 ft. that a 1,250,000-cir.mil cable has a resistance of 0.008282 ohm per 1,000 ft. and this would therefore be the required size. The total resistance would

2X0.004329 = 0.008658 ohm per

be slightly less than that specified, and we might try whether it would be possible to use one 1,250,000-cir.mil and one 1,000,000-cir.mil cable. The resistance of the latter size is 0.010353 ohm per 1,000 ft. and the combined resistance of the two would be their product divided by their sum, or 04601

ohm

per

1 ' 000 ft "

which proves to be to

consequently two 1,250,000-cir.mil cables would be used. As already found, the generator would be delivering 84,000 watts

great;

the lamps, which is

1,000 kw., the

is

equal to 84 kw.

to

Since the capacity of the generator -

lamp load would be

of its capacity, or 8. 4 per cent.

A .ULHJ

The voltage at the generator would be the voltage at the lamps plus the volts drop in line, or 120+12 = 132 volts. The watts lost in the line would be equal to the line current, 700 amperes, multiplied by the volts

= 8,400 watts loss. line, 12 volts, which gives 700X12 The load supplied has been found to be 84,000 watts; the line loss is 8,400 watts. To find what percentage of load supplied the line loss amounts to, we divide the line loss by the load and multiply by 100, drop hi

which gives -

X 100 = 10 per cent.

CHAPTER X

FUNDAMENTAL PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY Electromagnetic Induction. So far we have considered only the production of an electromotive force as generated in a battery cell by electrochemical action and by thermal action on the junction of two dissimilar metals. The most important method, and the one upon which the successful

application of electricity to-day depends is what is called electromagnetic induction the principle discovered in 183^ by Michael Faraday at the Royal Institution, London,

England

when

it is

that an electric pressure is set up in a conductor across the field of influence of a magnet.

moved

FIG. 133.

Conductor in a magnetic

field.

This principle is illustrated in Fig. 133, where a magnet shown with a conductor located in its field of force. If the conductor is moved upward or downward across the lines

is

of force at right angles to them, an electric pressure will be up in the conductor, and if its two ends are closed

set

133

ELECTRICAL MACHINERY

134

through another conductor, an electric current will flow. It does not make any difference whether the conductor or the

magnet is moved as long as there is a relative movement of the conductor through the magnetic field not parallel to the lines of force. shows the general idea embodied in the first used dynamo by Faraday. The device consists of a copper disk supported so that it could be revolved between the poles Fig. 134

FIG. 134.

Illustrates principle

Faraday's

first

dynamo.

of a horseshoe magnet. Brushes rest on the shaft and the periphery of the disk, so that when the disk is revolved a current of electricity flows through the circuit. This is the

principle upon which all dynamo-electric machinery is constructed at the present time. It has been ninety years since Faraday made this discovery, that when a conductor is

moved

across a magnetic field an electromotive force is generated in the former, yet with all the effort that has been given to finding out why this is so by hundreds of scientists,

PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY

135

still remains unanswered. Although we do not know why an electromotive force is produced in a conductor when it is moved across a magnetic field at right angles to the lines of force we understand the laws that govern the

the question

generation of an electric pressure by this means and are therefore able to design very efficient electrical machinery. Rule for Determining Direction of Electromotive Force. The direction of the electromotive force as generated in a conductor cutting the field of a magnet depends upon the direction of the lines of force

FIG. 135

and the direction and motion

'

FIGS. 135 and 136.

Eight-hand rule for determining direction of voltage generated in a conductor cutting the field of a magnet.

of the conductor.

There are a number of rules for determin-

ing the direction of the voltage when the direction of the lines of force and that of the conductor are known, but the

one shown in Fig. 135 is about the simplest, most convenient chiefly used. This method consists of placing the thumb first two fingers of the right hand at right angles to one

and and

another, as shown in the figure. Then if the forefinger points to the direction of the lines of force and the thumb the direc-

which the conductor is moved, the middle finger will indicate the direction of the voltage or current. In the figure the direction of current flow or voltage is away from the tion in

reader.

If the motion of the conductor is reversed as in Fig.

136, the direction of the voltage or current will be reversed.

The same would be true

if

the motion of the conductor were

ELECTRICAL MACHINERY

136

the same as in Fig. 135, but the polarity of the magnet changed; that is, the direction of the lines of force changed. If both the direction of the motion of the conductor

and the

lines of force are changed, the direction of the electromotive force will not be changed. For example, taking Fig. 136 as

the original condition, then Fig. 137 shows the condition when the direction of both the conductor and lines of force

have been changed. It is seen that the forefinger points in the same direction in both cases, or in other words, the direction of the voltage is the

Fig. 137.

Same

this the

deduction

same under both conditions.

From

as Fig. 135, except that both the direction of the conductor and lines of force are reversed.

may

be

made

that in order to reverse the

direction of the electromotive force generated in a conductor, it is necessary to change the direction of motion of either the

conductor or the lines of force, but not both. Lines of Magnetic Force. In Chapter II it was pointed out that the term "lines of force" is only figurative, because in the true sense, or at least as far as we know, such a thing as a definite line does not exist. Therefore, the lines which

are generally shown in pictures of magnets are used only to to indicate that there is a flow of "something" from the

N

magnet and the direction of the flow. If the like poles of two magnets are placed one centimeter, approximately one-third inch apart, as shown in Fig. 138, and they repel each other with a force of one dyne, the magthe

S pole

of the

PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY netic field

is

This unit

said to be of unit strength.

as the centimeter-gram-second (c.g.s.) unit

and

is

is

137

known

equivalent

1

to

445000

pounds.

Therefore, one line of magnetic force is just what the term would imply; namely, a force, just as one fan blowing against another creates a force which would tend to separate

them.

To express the magnetic unit more accurately, when

two magnets of equal strength, of 1

sq.

Shows a repulsion between two

FIG. 138.

cm. cross-section each,

like

magnetic poles.

are placed 1 cm. apart in air and repel each other with a force of one dyne, they are said to be of unit strength.

Unit Electromotive Force.

Taking a magnet having a

one line of force per square centimeter, and moving a conductor through this field at a uniform rate so that it will cut across it in one second, there field

of unit strength, that

is,

would be generated in the conductor what In other words,

electromotive force.

if

is

called a unit

a conductor cuts

one line of force per second, it will generate a voltage of unit value. This unit has never been given any other name than unit electromotive force. It is a theoretical unit and is used only in establishing the relation between trical

and magnetic

The value

it

and other

elec-

theoretical units.

of this theoretical unit of voltage

is

so small that

it

would

and

to get a unit for use in practice 100,000,000 of these theoretical units are combined into one and called a volt. Expressing this another way, when a conductor cuts 100,000,000

be very inconvenient to use

in practice,

per second, at a uniform rate, pressure generated in it.

lines of force

Right here

is

lines of force cut

it

will

have one volt

of electric

where we should distinguish between the number of and the rate at which they are cut. It is not necessary

138

ELECTRICAL MACHINERY

for a conductor to

move

for

one second and cut 100,000,000 lines of

force at a uniform rate in that time, in order that one volt be generated in it. If such were the case, it would have one volt generated in it during

On the other hand, if the conductor was in motion for only one-hundredth of a second and cut 1,000,000 lines of force, it would have generated in it one volt, the same as when cutting 100,000,000 lines of force in one second the rate of cutting is the same in both cases. From this it is seen that all that is necessary to generate one volt is for the conductor to cut the lines of force at a rate equivalent to 100,000,000 per second. This might be called the starting point in the the period of one second.

design of

all

dynamo-electric machinery.

Fundamental Principle of the Electric Motor.

Before

giving further consideration to the electric generator, it is necessary that the fundamental principle of the electric

motor be understood. electric current flows

We have already seen that when an through a conductor it sets up a mag-

netic field about the latter, as in Fig. 139, and that when a conductor moves across a magnetic field so as to cut the lines of force, an electromotive force is generated in the conductor, which will cause a current to flow if the ends of the con-

ductor are connected together. This brings up the question what effect a magnetic field has upon a conductor carrying a current when the latter is placed in the former. Fig. of

140 shows a horseshoe magnet, the magnetic field of which will be, if not subjected to an influence, uniformly distributed as indicated. If a conductor through which a current is flowing is brought near the field of a magnet, as in Fig. 141, the result will be that shown in the figure. If the current is

flowing up through the plane of the paper, the lines of force about the conductor adjacent to the field of the magnet are

same direction as those of the latter. This being the two fields are forced in together between the conductor and the magnet and cause a distortion of both as shown. Lines of magnetic force always act like elastic bands in the

case, the

in tension; they may be distorted, but tend to take the shortest path, and that is just what is illustrated in Fig. 141. Under the conditions shown, the field of force of the magnet

tends to push the

field of the

the conductor to the

left,

conductor and along with by the arrow.

as indicated

it

PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY If the field,

from

conductor

is

139

placed in the center of the magnetic

as in Fig. 142, the result will be as given. The flux the magnet and about the conductor are in the same

direction on the right-hand side of the latter, but opposed on Consequently, part of the flux from the

the left-hand side.

magnet on the left-hand

side of the conductor will be dis-

torted to the right side, as shown, causing even a greater distortion of the field than in Fig. 141, and will also tend to

cause the conductor to

move

to the left.

Illustrate the

just

FIG. 142

FIG. 141

FIGS. 139 to 142.

What we have

fundamental principle of the motor.

the fundamental principle of the 'electric motor and be expressed thus: A conductor carrying a current in the field of a magnet tends to move to cut the lines of force of the latter at right angles. This, it will be seen, is the

seen

is

may

inverse of the fundamental principle of the electrical generator, which is, a conductor moved across a magnetic field so as to cut the lines of force has

duced in

an electromotive force

in-

it.

Direction

Conductor Will Move in a Magnetic Field

Depends Upon the Direction of the Current. It is evident from what we have seen in Figs. 141 and 142 that the direction the conductor will tend to move in depends upon the

ELECTRICAL MACHINERY

140

direction of the current in the conductor

and the

lines of

force of the magnetic field. The direction can be determined in the same way as for the direction of the electromotive force developed in a conductor, but using the left hand Placing the first three fingers of the left hand at right angles to each other, as in Fig. 143, if the

instead of the right.

middle finger points in the direction of the current in the conductor and the forefinger in the direction of the lines of force, then the thumb will always point in the direction that

FIG. 143

FIG. 143.

FIG. 144.

Left-hand rule for a motor. Right-hand rule for generator.

the conductor will tend to

move

in.

Hence,

if

either the

direction of the current or the lines of force are reversed, the direction of motion of the conductor will be reversed. This is

true in all cases,

when

the rotational direction of a direct-

current motor's armature

is reversed; the direction of the reversed through the armature conductors or though the field coils, the latter changing the direction of the

current

is

magnetism of the

field poles.

Counter-Electromotive Force. dicated, the conductor will

In Fig. 142

and current tend to move

that with the lines of force

we have

seen

in the direction into the left.

As soon

PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY as the conductor begins to move,

it

141

will be cutting the field

of the magnet. Consequently, an electromotive force will be generated in the former. By applying the rule, Fig. 144, for the direction of the electromotive force in the conductor, it will

be

seen that

the

generated electromotive

opposite to the direction of the current

force

is

is, the voltage in the the conductor is to developed opposite pressure that in This back presthe to flow the conductor. causes current ;

that

be termed, is called a counter-electromotive force and, as will be seen in Chapter XVI, is an important factor in the operation of the electric motor. What has been sure, as

it

may

shown

in this and in Chapter XII indicates that whenever a conductor moves in a magnetic field or a magnetic field is moved across a conductor so that lines of force are cut, no matter what the conditions may be, there will always be a

voltage induced in the conductor. From this very fact many of the most serious problems in the operation of electrical

machinery and circuits arise. How an Electric Generator is Loaded. In Fig. 145, if the conductor is moved upward, as indicated, a voltage will be induced in it that will cause a current to flow in the direction shown. However, from what we have learned in the foregoing, when the current flows in the conductor it will tend to cause the latter to

move

in a direction to cut the lines of

force or, in other words, produce the condition of an electric motor. This is just what would happen in Fig. 145. Mechan-

applied to cause the conductor to move upward, which causes the current to flow in the circuit. The magnetic field set up by the current in turn reacts upon the flux from the magnet and also tends to impart motion to the conductor. By applying the rule for the direction of motion of the conductor due to the current it is seen that it tends to move downward, or opposite to the direction the mechanical ical force is

This point brings out the way is causing it to move in. that an electric generator is loaded. The source of driving power turns the armature in a given direction, which proforce

duces a voltage and causes a current to flow; the current

ELECTRICAL MACHINERY

142

flowing through the armature conductors in turn produces a reverse, or counter-turning effort to that produced by the steam engine or other source of motive power driving the generator. However, the counter-turning effort caused by the action of the current on the magnetism from the poles cannot be so great as the applied power, or the machine

would

stop. But the greater the number of amperes supplied the armature for a given voltage, the greater the counterby turning effort. Consequently, the source of motive power

will

have to develop a greater

FIG. 145.

Illustrates

how

current

effort to

is

overcome that pro-

caused to flow in a

circuit.

duced by the current. The same way with a motor, the counter-electromotive force cannot be so great as the applied voltage, or no current would flow through the conductors on

and the machine would stop. The foregoing shows that the electric motor and the generator are very In fact, the fundamental principle of closely interrelated. one is always embodied in the other. The machine that can be used for a motor can also be used for a generator, the only

the armature

difference in their design being slight modifications to obtain certain desired characteristics.

Simplest

Form

of Electric Machine.

The windings on

the armature of a generator consist of a series of loops or coils grouped in various ways, depending upon the type of

PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY

143

machine, voltage, etc. The simplest form would be one loop arranged to revolve between the north and the south pole The ends of the loop of the magnet as shown in Fig. 146. connect to the rings R : and R 2 with brushes B^ and JB, 2 resting on the latter to form a rubbing contact between the Conrevolving loop and the stationary external circuit C. ,

sidering the loop to revolve in a clockwise direction, as indicated by the curved arrow, the side of the loop under the pole will be moving downward while the side under the S pole

N

will be

moving upward.

The

lines of force are

from the

N

to

the S pole therefore, by applying the rule for the direction of the electromotive force generated in a conductor cutting lines of force, it will be found to be as given by the arrows on ;

the two sides of the loop, which is away from the reader under the pole and toward the reader under the S pole. This is

N

just as it should be, since the lines of force are in the same direction under each pole, but the direction of the conductor

under one pole is opposite to that under the other. Factors Governing Value of the Voltage. By tracing around through the loop it will be seen that the e.m.f. generated in the side under one pole is added to that under the other pole. Or, in other words, we have the same condition as when two voltaic cells are connected in series, and if two volts are generated in one conductor, the two conductors in series will generate four volts.

Hence,

it is

seen that one of

the factors which govern the voltage of *a given generator would be the number of conductors connected in series. For

instead of only one turn in the

coil, as in Fig. 146, in series, as in Fig. 147, and if the coil is revolved at the same rate and the magnetic density the same in both cases, then each conductor under a pole will have

example,

if

we have two turns

equal voltage generated in it. Again, by tracing through the be seen that four conductors are in series; con-

coil, it will

sequently the voltage generated in the coil will be four times that in one conductor, or in other words the voltage increases as the

number

Another way

of turns in the coil

to increase the voltage

is

increased.

would be

to increase

144

ELECTRICAL MACHINERY

FIG. 146

Fia. 148

FIGS. 146 to 149.

FIG. 147

FIG. 149

Single-coil alternating-current generators.

PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY

145

the number of revolutions per was doubled, the number of lines of force cut by each conductor would be doubled. Consequently, the voltage would be increased by two. A third the speed of the coil that minute made by the coil ;

is, if

that the voltage generated in the coil may be varied is by changing the number of lines of force in the magnetic field. If the speed of the coil remains constant, but the strength of

way

the magnetic field is doubled, then double the number of lines of force will be cut in a given time. The latter is the one way usually employed for varying the voltage of all

modern generators and

is treated in Chapter XIII. Current Reverses in External Circuit. In Fig. 146 the flow of the current is from conductor a to ring R 2 and brush B 2 through the external circuit C and back to brush B i and ring R l and back into conductor &, thus completing the circuit.

When

made

the coil has

one-half revolution, as shown in Fig. under the pole and conductor ft

148, conductor a will be

under the S

N

shown, with the result that the direction of the voltage generated in the two conductors is reversed. The direction of the e.m.f. in conductor a, Fig. 146, is toward pole, as

the reader, but in Fig. 148 it is away in b, Fig. 146, the direction is away from, while in Fig. 148 it is toward the reader. The result of this change in direction of the' voltage in the coil is a change in direction of the current in the external ;

by the arrowheads.

circuit, as indicated

From

this it will be

seen that on one-half of the revolution the current

is

flowing

through the circuit in an opposite direction to that on the other half of the revolution; that is, the current is caused to flow back and forth through the circuit. If the voltage in the armature conductors change in direction as they pass alternate north and south poles there must be some position where the voltage is zero When the coil

moving

;

this is indicated in Fig. 149.

in the position shown in Fig. 149, it is parallel with the lines of force and is therefore not is

cutting them, and consequently not producing any voltage. From this point the voltage increases until the conductors are at the center of the polepieces,

where they are moving

at right

ELECTRICAL MACHINERY

146

angles to the lines of force and are therefore cutting the flux maximum rate, consequently producing a maximum pres-

at a

sure.

For the next quarter

of a revolution the voltage de-

creases to zero.

Electromotive Force or Current Curve.

The

series of

values that the voltage or current passes through in the coil during one revolution may be expressed in the form of a curve, Fig. 152.

The distance along

FIG. 151

FIG. 150

FIGS. 150 and 151.

the straight line between the two

Single-coil direct-current generators.

zero points of one curve represents the time required by the coil to pass the pole faces, or in Figs. 146 to 149 to make one-

half revolution.

The

vertical distance between the line

and

the curve at any point represents the value of the voltage or current in the coil at that instance. The curve above the line

represents current or voltage in one direction, while the curve below the line represents current or voltage in the opposite direction. current or electromotive force that changes in

A

shown in the foregoing force. or electromotive current alternating direction in the circuit as

is

called

an

PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY

147

The voltage generated in the armature of all commercial types of generators is alternating, no matter whether the current in the external circuit flows in one direction or is alternating back and forth. If we want the current to flow in one direction in the external circuit, or, as it is usually called, a direct current or continuous current, some means

must be provided

to

change the alternating voltage generated

Maximum l/a/i/6

Maximum l/a/u&FIG. 152.

Alternating-current or voltage curve.

in the armature coils to one that

is

always in the same direc-

tion in the external circuit.

In Fig. 150 is shown a scheme that will maintain the voltage in one direction in the external circuit. Instead of the ends of the coils connecting to two rings, as in Figs. 146 to 149,

In the coil on segment 8 1 and the current

they connect to the two halves of a divided ring.

B

position shown, brush in the external circuit

is

t

rests

in the direction indicated.

When

the

has revolved one half revolution as in Fig. 151, brush B : is resting on segment S 2 and although the current has reversed

coil

,

150, it is maintained in the same direction in the external circuit, as indicated by the arrowheads.

in the coil

from that in Fig.

Although the voltage is applied in one direction to the external circuit, the current will not be of a constant value on Maximum Value

FIG. 153.

Maximum Va/ue

Pulsating-current or voltage curve.

account of the varying value of the voltage. What will be obtained is a current that flows in waves, as shown in Fig. 153

ELECTRICAL MACHINERY

148

is known as a pulsating current. To obtain a constant current for a given value of resistance in the external circuit, or, as it is usually called, a direct current, it is necessary to have

and

a number of as

many

coils

on the armature and the ring divided into

sections as there are coils.

Machine with Ring-Type Armature. So far we have only considered dynamos that have one coil of a single turn of wire on the armature.

FIG. 154.

In Fig. 154

is

shown, diagrammatically,

Diagrammatic representation of ring-armature type generator.

the complete layout of a dynamo-electric machine having an armature of the ring type. Coils of wire, designated field The winding on the coils, are placed on the polepieces.

armature

shown, for simplicity's sake, to be continuous for the entire circumference of the core and closed on itself, with a tap taken out at each turn of the winding to a bar or segment in the commutator. This winding could have been shown is

grouped into coils, as in Fig. 155, with the leads of each coil coming out to two commutator bars, as shown, which is generally the way the job is done in practice, but for our purposes

PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY

149

Fig. 154 is better suited. The winding in Fig. 155 has twice as many turns as that iji Fig. 154, consequently, will generate twice the voltage under a given condition of speed and field strength.

In Fig. 154,

if

a current

is

coils in the direction

it

caused to flow through the field will cause the top polepiece to

shown, become south and the bottom one north polarity, and the magnetic flux will flow in the direction indicated.

FIG. 155.

Same

Then,

if

the

as Fig. 154, but with two turns per coil on the armature.

armature

is revolved in the direction of the curved arrow, the conductors under the S pole will be cutting the lines of force in a left-hand direction, while those under the N pole will be

cutting the flux in a right-hand direction. The lines of force to the S pole, in each case, are moving upward, from the the rule for therefore, by applying determining the direction

N

of electromotive force,

it

will be

under the S pole the voltage

found that in the conductors

down through

the plane of the in while the conductors under the paper, pole it is toward the reader, as indicated by the arrowheads. It will be seen that is

N

ELECTRICAL MACHINERY

150 all

the voltages generated in the various conductors under the series assisting one another, likewise under the

S pole are in

N

pole. Therefore, the sum of the voltages generated in the conductors under one pole is the voltage that will appear at

when

in the position shown. It will windings under the pole opposes that generated in the conductors under the S pole. Since the voltages in the two halves of the windings are equal,

the brushes

the latter

is

N

also be seen that the voltage in the

or at least should be, no current flows in the winding as long is open between the two brushes.

as the circuit

The question of electromotive-force generation was discussed to considerable extent in Chapter III, and it was pointed out that the device which supplies the current to the circuit does not generate the current, but produces a voltage

that causes the current to flow in a conductor

when

it is

con-

nected between the positive and negative terminals of the source of voltage. This is just the condition we have in Fig.

The conductor on each half of the armature cuts the line of force and generates a voltage; however, the arrangement of the winding is such that the voltage in one half opposes that in the other half, and no current can flow around 154.

winding until the brushes are connected to an external In the figure the external circuit is shown on the center of the commutator; however, this is for simplicity's sake only, as this circuit might be a motor or a group of lamps, or any device that requires an electric current for its operation, and might be located at a considerable distance from the machine. A generator in an electric circuit is the same as a pump is in a circulating system. The pump does not generate the fluid

in the

circuit 0, as in Fig. 156.

it causes to flow in the system, but creates a pressure that causes the fluid to flow in the pipe line. Likewise an electric generator only produces the pressure that causes the electric

that

current to flow in the circuit.

The two halves of the armature windings in Figs. 154, 155 and 156 are similar to two voltaic cells in parallel. In Fig. will be seen that the electric pressure generated in the conductors under the pole causes a current to flow out from

156

it

N

PRINCIPLES OF DYNAMO-ELECTRIC MACHINERY

151

the positive brush through the external circuit C and into the negative brush, as indicated by the arrowhead; likewise, for the conductors under the S pole. Since the two halves of the windings are in parallel, the voltage appearing at the brushes

one half of the winding, just as when two voltaic cells are connected in parallel the voltage of the group is equal to that of a single cell. Also, each half of the will be that developed in

will

winding circuit

;

that

Same

FIG. 156.

supply one-half of the current in the external when the armature is supplying 30 amperes to

is,

,is

machine in Fig. 154, but with brushes connected to an external circuit.

the external circuit

ductors under the

under the S

pole.

voltaic

are

cells

current to a circuit

each

(7,

N

15 amperes will be flowing in the conand 15 amperes in the conductors

This

is

again the same as

when two

parallel and supplying one-half of the current is supplied by

connected

in

cell.

Although the foregoing discussion has been in reference to two-pole machines, the general principle applies equally to multipole machines, although the division of the current may be somewhat different in the armature windings.

CHAPTER XI DIRECT-CURRENT MACHINERY CONSTRUCTION Types of Armatures.

Direct-current armatures

may

be

classed under two

Both general types ring and drum. types get their names from the shape of the core. The core of the ring type consists of an iron ring about which the coils are wound, as shown in section in Fig. 157, where the coils on a drum armature are placed on the surface of the core, or in slots in the surface of the core, as in Fig. 158.

Fig. 158

Cat Leads;

FIG. 157.

King armature

in section.

shows three coils in place and how the coils fall over one another to form a complete winding, as in Fig. 159. In the drum-type armature the spread of the coils that is, the number of slots spanned by a coil is determined by the number of poles. For example, in Fig. 158 the coils span approximately one-quarter of the core, which would indicate that this armature is intended to operate in a four-pole field frame. In the ring armature the distribution of the winding on the core is the same, irrespective of the number of poles, '

152

DIRECT-CURRENT MACHINERY CONSTRUCTION

153

where in the drum armature the coil spans approximately the distance between the centers of adjacent poles. Although there is no difference in the two types of armatures so far as voltage generation is concerned, when it comes to a consideration of the various elements that take

place in the windings, the ring type lends itself

FIG. 158.

Drum-type armature core with three

much more

coils in place.

readily to a theoretical discussion, therefore consideration of this subject.

is

used in our

In the commercial type of machines the armature connumber of coils. On the small-sized machines the

tains a

is usually wound with a small number of coils a considerable number of turns of small wire, whereas having in the large-sized machines the armature is wound with a

armature

FIG. 159.

Drum-type armature complete.

large number of coils of large wire having a small of turns, usually one turn made from a copper bar.

Objections to Ring-Type Armatures.

number

In the earlier type

of machines the ring-armature construction was used to considerable extent, but it has since been practically abandoned.

Some

of the objections to this type of construction are that only one side of the coil is effective in generating voltage.

ELECTRICAL MACHINERY

154

Why ture

this is so is explained in Fig. 160,

where a ring arma-

shown between the

poles of a two-pole frame. It will be seen that all the lines of force are only cut by the conis

ductors on the outer surface of the core.

Therefore, only these parts of the coils are effective in generating voltage. There is always a leak across the space in the center of the ring; that is, a small percentage of the lines of force, instead of flowing around through the core, take the path

FIG. 16 J.

Shows magnetic leakage

in ring

armature.

across the space in the center of the ring, as indicated in Fig. 160. The conductors on the inside of the ring cut the lines of force that leak across from one side of the ring to the other, in the same direction as the conductors on the outside of the core. Consequently, a voltage will be induced, in the same direction, in the side of the coil on the inner

periphery of the ring, as in that on the outer. In Fig. 160 consider the ring revolving in the direction of the curved arrow; then under the N pole the voltage in both sides of the coils

is

up through the plane of the paper, and

DIRECT-CURRENT MACHINERY CONSTRUCTION under the S pole

it is

away from

the reader.

155

In either case

evident that the voltages generated in the conductors on the outside and inside of the ring oppose each other. Since only a small percentage of the flux leaks across the ring, it is

only this percentage will be cut by the conductors on the inner periphery, and the voltage generated in these conductors will be only a small percentage of that in the outside, the difference between the two being the effective voltage in the coil. The foregoing is another objection to the use of a

ring armature.

Another difficulty is in winding the coils on the core, they have to be wound in place by hand. On account of having to thread the coils through the center of the core, the placing of the winding is a somewhat long and tedious job. These and other structural and electrical defects have caused this type of construction to be practically abandoned in favor of the drum type of armature.

Drum-Type Armatures. The core of the early types of drum armatures consisted of a cast-iron cylinder keyed on a shaft, as in Fig. 161.

Pieces of fiber were placed in small

f-lrvn Core

.-Fiber

FIG.

161.

Smooth

core for drum-type armature.

slots in the corner of the core, as shown, to facilitate the spacing of the coils around the periphery. One of the serious objections to the use of solid cast-iron cores was that they

had heavy current generated

in them, which not only greatly increased the temperature for a given load, but also loaded up the machine.

ELECTRICAL MACHINERY

156

The foregoing will be understood by considering Fig. 162, which shows an iron core between the N and S poles of a magnet. If the cylinder is revolved in the direction of the curved arrow, then the side of the core under the N pole will be cutting lines of force in a right-hand direction and will have a voltage induced in it that will tend to cause a current

toward the reader. On the other hand, the side of the cylinder under the S pole is cutting the flux entering the pole in a left-hand direction, and consequently has a voltage to flow

induced in

it

that will tend to cause current to flow

from the reader.

FIG.

162.

This

is

just

away what we found out about a

Shows how eddy currents are generated

in

armature core.

loop of wire revolved between the poles of a magnet in

Chapter X.

The condition in Fig. 162 is such that the voltage generated in one side of the core is assisting that in the other side. Consequently, a current will flow around in the core, as indicated by the dotted loop and arrowheads. This current is entirely independent of the winding on the armature and

external circuit, and just as long as the field poles are and the armature revolved, a current will circulate

excited

or eddy around in the core. Since these currents circulate or eddy around in the core as water in a whirlpool, they are called

eddy currents.

Eddy Currents. These eddy currents represent a distinct loss, not only in capacity, but also in the power used to drive the generator or motor. Eddy currents Effects of

increase

the

temperature

of

the

machine,

consequently

DIRECT-CURRENT MACHINERY CONSTRUCTION

157

reduce the useful temperature range. For example, if the normal no-load temperature of the armature, if eddy currents did not exist, is 80 deg. F., and the maximum temperature that the machine can be operated at is 212 deg. F., then the machine can be loaded to the extent that would

increase the temperature from 80 to 212 deg. F., or 132 deg. But on the other hand, suppose that the no-load temperature of the armature, due to eddy currents, is increased to 100 deg. F. then the machine is only capable of carrying a use;

ful load that will increase the temperature

from 100

to 212

deg. F., or 112 deg. Consequently, the useful capacity of the machine under the latter conditions will be reduced. It should be kept in mind that the amount of load that can be carried by any electrical machine is limited by the heating effect of the load. For a machine insulated with fibrous material this temperature must be limited to about 212 deg. F. Another effect of eddy currents in the armature is to increase the power necessary to drive the machine. This will be understood by referring to Fig. 162. Here the generated current due to the core revolving in the magnetic field in the direction indicated by the curved arrow is

indicated by the dotted loop and arrowhead in the core. Current flowing in the magnetic field will cause a pull to be exerted upon the core, just as explained for a single con-

ductor carrying a current in a magnetic

field in

Chapter X.

The direction

of the pull on the core may be determined for rule the the direction of a motor, or, in Fig. 162, it by will be found that the eddy currents in the core will produce

a pull against the direction of rotation. In other words, we assume that the core is revolving in the direction of the

curved arrow, but the direction of the eddy currents in the armature core produces a pull in the opposite direction. Consequently, the source from which the armature is driven will have to develop power enough not only to drive the armature to supply its useful load, but also to overcome the effect of the

eddy currents

in the core.

ELECTRICAL MACHINERY

158

Eliminating Eddy Currents. From what we have just it is apparent that/ if possible, these eddy currents should be eliminated. This is done to a very large degree by building up the armature core of thin sheets of soft iron or These sheets are from 0.01 to 0.03 in. steel, as in Fig. 163. in thickness. In the early type of machines the oxide on the seen,

was very largely depended on

surface of the sheets sulate one

from the

insulate the disks

other.

Although from each other,

to in-

this did not completely it

offered considerable

resistance to the flow of the current in the core parallel with the shaft. In the modern machines the iron sheets that the

core is made of are given a very thin coat of insulating varnish on one side, which practically entirely eliminates the effect of

eddy current. Cast-Iron-

FIG. 163.

\t."L am/noted Iron

-j

Shows how armature cores are

built

up of thin sheets of

iron.

account of the insulation the core has to be made slightly longer than a solid core would be, in order to get in

On

Armature cores that are

the same volume of metal.

built

up

of thin sheets of iron are said to be laminated, and the sheets are frequently referred to as the laminae. The cores of smallsized armatures are

to the shaft

keyed

and held between

cast-iron shrouds, or retaining plates, which are held in place by a nut threaded on the shaft, as in Fig. 163, or by bolts run

through the

core.

Objection to Smooth-Core Armatures. With these smooth-core armatures, the winding had to be placed on the surface of the core. There were several objections to this,

such as the coils, being on the surface of the core, were exposed to mechanical injury sufficient space must be allowed between the armature core and polepieces for the winding. ;

DIRECT-CURRENT MACHINERY CONSTRUCTION

159

On

account of the comparatively long space between the armature core and polepieces, considerably more power is required to be expended in the field coils to cause the lines of force to flow from the latter to the former, or vice versa,

than

if

the core

was

as near the field poles as mechanical construction.

would be con-

with good Another serious objection is that the coils are wound on the core by hand, one coil at a time. Consequently, the coils can only be removed the reverse of the way they are put on. Therefore, if two or three coils are injured in the winding, it generally means that the whole winding must be removed and a new one put in its place, whereas, if the coils are sistent

made up

separately, as in

FIG. 164.

modern machines, the injured

Slotted core for drum-type armature.

can generally be removed and replaced by new ones, by removing only a small part of the total winding. All the difficulties cited are practically eliminated by slotting the core as in Fig. 164. The core is built up of thin sheets

coils

as in Fig. 163, but instead of the outer periphery of the disk being smooth as in the figure, it has slots cut in it. These

some of which are shown in Figs. However, when the coils are made up and insulated before they are put on the core, the slots in the core must be open at the top, as in Figs. 166 or 167. Where the coils are made up separately and insulated before placing on the armature, it is evident that the winding is not only more easily put in place, but can be better insulated. The placing of the coils in slots in the armature core protects them from mechanical injury in case the bearings slots take different forms,

165 to 167.

ELECTRICAL MACHINERY

160

wear and allows the armature to rub on the polepieces; also allows the space between the core and the field poles be reduced to a

minimum

Fia. 165

construction.

a

consistent with good mechanical

Fia. 166

FIGS. 165 to 167.

FIG. 167

Different shapes of armature-core slots

Consequently, the magnetic

minimum power

it

to

expenditure in the

field is set

field

up with In the

coils.

larger-sized armatures a cast-iron spider, as it is called, is keyed to the shaft and the laminated core built upon the spider, as in Fig. 168.

Laminated Iron Core.

Casrlron.--

Spder

FIG. 168.

Core for large armature built upon a cast-iron spider.

Field Magnets.

The function of the field magnets in an motor is to furnish the magnetic field,

electric generator or

which in a generator is cut by the armature conductors to generate voltage, and in a motor reacts upon the current flowing in the armature conductors to produce rotation. In the development of the dynamo-electric machine the field magnets have taken on a multiplicity of forms. The field magnets of the earlier types of dynamos were permanent horseshoe magnets, similar to that shown in Fig. 169. to-day this type of field pole

is

used, in

some

cases,

Even

on small

DIRECT-CURRENT MACHINERY CONSTRUCTION

161

and other purposes. Howmagnet was never used on machines of any considerable size, chiefly because the magnets would have to be very large the strength of the magnets decreases when in use, owing to the vibration of the machine and the effects of the magnetic field set up by the current in the armature

magnetos for

ignition, signaling

ever, this form, of

;

is no way of controlling the of is which the chief means usually emthe strength field, for the ployed controlling voltage of the generator or the

winding; also because there

FIG. 169

.

BRASS FIG. 170

Horseshoe magnet field pole. Edison two-pole dynamo, or motor.

FIG. 169. FIG. 170.

speed of an adjustable-speed motor. These defects soon led to the adoption of electromagnets; that is, coils of wire placed on polepieces of soft iron and excited from some source of electric current.

Early Types of Field Magnets. Since the permanentfield poles were of the horseshoe shape, it is to be expected that most all of the earlier electromagnets used for field poles were of this form. Fig. 170 shows one of the and Fig. 171 is a Edison of machines, two-pole early types somewhat later and improved type of the same machine. In this arrangement of poles, if they were mounted on an iron

magnet

162

.

base

it

would

ELECTRICAL MACHINERY

short-circuit the magnetic field

;

that

is,

instead

N

of the lines of force passing from the pole across the air gap, and through the armature core into the S pole, they

would take the easier path around through the iron base. To overcome this defect, a nonmagnetic plate of brass or zinc was placed between the polepieces and the baseplate as To prevent the lines of force from leaking out indicated. along the armature shaft

down through

the bearing pedestals

FIELD COtL$

ZINC OR BRASS-"

PLATE

FIG. 171.

Improved type of Edison two-pole dynamo or motor.

into the base of the

machine and back into the polepieces,

the pedestals were usually made of brass. To get away from the nonmagnetic bearing pedestal and bedplate, the polepieces were turned upside down with the

armature placed in the top, as in Fig. 172. With this arrangefield magnetism passes from the N pole into the armature through the armature' core into the S pole, and down around through the baseplate. Another form of field

ment the

magnet is that in Fig. 173. This type was usually mounted on a wooden base, for the same reason that the nonmagnetic plate was used in Figs. 170 and 171.

DIRECT-CURRENT MACHINERY CONSTRUCTION

163

the foregoing schemes the flux from the polepieces passes directly from one field pole into the armature, and then to the opposite pole and around through the field

In

all

Such an arrangement

structure.

is

called a salient-pole

machine.

Fia. 172.

Machine with the

field

poles reversed from that in Fig. 171.

Consequent-Pole Type Machine. Another type of field pole used in the development of the electric machine is given in Fig. 175. In this construction if the top of one field coil is

made north and

from the

N

the other south, the lines of force will flow to the S pole without ever passing ail. To overcome this difficulty the

pole around the armature at through

top ends of both coils are made the same polarity; therefore the bottom ends must also be the same polarity, as shown in the figure. other,

In this arrangement the two N poles oppose each lines of force must take the next easiest path,

and the

ELECTRICAL MACHINERY

164

which is down through the armature to the S pole. A machine having a field frame in which like poles oppose each other, so as to cause the flux to pass through the armature, is called a consequent-pole machine. One of the serious objections to this type is that the opposing poles cause a heavy

magnetic leak around through the

air

from the

N

to the S

N

pole; that is, instead of all of the flux passing from the pole into the armature and then to the S pole, a large number of the lines fly out in all directions into the air and around to the

S

pole.

This leak constitutes a direct

loss.

In

all

Wood

FIG. 174

Fia. 173

FIG. 173. FIG. 174.

Bipolar dynamo or motor on wooden base. Common type of four -pole dynamo or motor.

types of machines there is always a certain amount of magnetic leakage, but it is much more pronounced in the consequent-pole machine than in the salient-pole type. Motors or generators with only two poles are called bipole machines; those having more than two poles, that is, four, six, eight, etc., are called multipolar machines. None of the types of field frames so far considered lend themselves

readily to multipole construction, consequently very these types were developed into multipole designs.

few of

The arrangements of frame that have been exploited could be

Materials Used in Field Poles. poles in the field

carried out almost indefinitely, but the one design that is almost exclusively is the arrangement shown in

now used Fig. 174.

This construction

is

of the salient-pole type; that

DIRECT-CURRENT MACHINERY CONSTRUCTION is,

there are no opposing poles.

165

This design can be used as

readily for bipole as for multipole construction. All of the earlier field frames and polepieces were constructed of cast iron or steel. In some of the modern types

the whole field structure

is

thin sheets of iron or steel.

laminated; that is, built up of Others again have a cast-iron

yoke to which laminated pole-pieces are bolted. The yoke is the circular portion in Fig. 174. Other types have cast-iron polepieces with laminated poleshoes. The poleshoe is indicated in Fig. 174. This part is built up of thin sheets of iron or steel

and bolted

FIG. 175.

to the cast-iron polepiece.

Consequent-pole type dynamo or motor.

X

it was Function of the Commutator. In Chapter if a single coil is connected to rings R l and R 2 as in Fig. 176, and revolves between the poles of a magnet as indicated, an alternating current would be caused to flow in

shown that

,

It was also shown that this alternating current could be changed into a current that flows in one direction that is, a direct current by connecting the ends of This the coil to a divided ring 8 t and $ 2 as in Fig. 177.

the external circuit C.

,

divided ring represents the simplest form of what is called a commutator on a direct-current machine. Fig. 177 also represents the simplest form of .an armature, one that has only one coil

revolving in a two-.pole

field.

direct-current generators a large

In the commercial type of

number

of coils, depending

ELECTRICAL MACHINERY

166

upon the size of the machine, are arranged on the armature and connected to a commutator. Commutator Construction. The commutator is made up of a number of copper segments similar to the one in Fig. 178, each segment being insulated from the other. These segments are slotted in one end, as shown, so that the armature-coil leads may be easily connected. Many of the large-sized commutators

have an extension soldered to each segment or bar, as in Fig.

FIG. 177

FIG. 176.

Single-coil alternating-current generator. FIG. 177. Single-coil direct-current generator.

179, so that, instead of the coil leads being bent down to the commutator, they are brought out almost on a level with the periphery of the armature. Fig. 180 shows a section through a common type of com-

mutator.

The black

lines

A

across the surface are insulation,

usually mica, between the segments. The heavy black lines B are also insulation, therefore it is evident that each segment is

insulated from the iron or steel form.

commutator

is

called a bar or segment.

Each division on the The bars and insula-

DIRECT-CURRENT MACHINERY CONSTRUCTION

167

and sleeve C. Then the front are insulation and its vee, D, put in place and the nut E, which is threaded on the sleeve (7, is screwed up as tight as it can be drawn. On account of the expansion and contraction tion are assembled on the vee

by wide variations of temperature

of the commutator, caused

and

due

it is necessary that the hold the bars and insulation clamping rings very tight, or there will be a movement of the bars that will cause trouble.

strains

to centrifugal force,

The insulation used between the bars is usually mica or 1 micanite, about / 32 in. in thickness, and in fact this is the material that has been found that will stand up under all only conditions. Micanite consists of thin flakes of mica built up into sheets and held together by a suitable binder. In some of the small-sized commutators the bars are molded into the

FIG. 179

FIG. 178

FIGS. 178 and 179.

Types of commutator bars.

metal sleeve with an insulating compound. This construction for small-sized machines seems to give satisfactory results. In the early development of the art various materials were tried for insulating commutators, such as red fiber, fish paper,

and various combinations of these materials and mica and other insulations built up in alternate layers, but all have been discarded. One of the troubles with most of the substitutes for mica insulation in commutators is, they are easily eaten away in asbestos, etc., also

case of sparking at the brushes. Furthermore, all fibers or or less more contraction and expansion to are subjected papers

due

to

moisture conditions.

ness between the bars

and

This eventually led to slight looseinsulation, so that oil or copper

and

ELECTRICAL MACHINERY

168

carbon dust could penetrate and cause pitting of the commutators. tion

is

that

One it

temperature

;

of the chief requisites of a commutator insulashall not be affected by moisture and changes of

also, it

must possess a certain

FIG. 182

FIGS. 180 to 183.

elasticity so that

it

Fia. 183 J|

Types of commutators.

the space between the bars irrespective of the expanof the commutator. So far, mica seems

will

fill

sion

and contraction

to possess these requirements to a greater extent than anything else. The story of commutator insulation is one of the most

interesting chapters in the history of electrical development.

DIRECT-CURRENT MACHINERY CONSTRUCTION

169

other kinds of commutator construction are used, especially in the older types of machines, besides that shown in Fig. 180, some of which are indicated in Figs. 181 and 182,

Many

from which

it is

seen that the general principle

is

the same.

In the larger-sized machines the commutators are built upon a cast-iron spider, the same as the armature core. Fig. 18,3 shows a cross-section of a large commutator that is representative of large-sized construction.

CHAPTER XII INDUCTANCE AND COMMUTATION Magnetic Field About a Conductor. It has already been when an electric current flows through a conthe former causes a magnetic field to be set up about ductor, the latter. This magnetic field takes the form of concentric The direction of the circles, as shown in Figs. 184 and 185. learned that

PIG. 185

Fia. 184

FIGS. 184 to 185.

magnetic

The rent

field is

Magnetic

field

about an electric conductor.

determined by the direction of the current.

cross in the center of the conductor indicates that the curis

assumed

to be flowing

this direction of current the

direction indicated

current

away from the magnetic

by the arrowheads.

reader,

and for

field revolves in the

If the direction of the

reversed as in Fig. 185, the direction of the magnetic field is reversed, as shown. is

How

Voltage Is Induced in a Conductor. This magnetic field seems to emanate from the center of the conductor and has an increasing diameter as the current increases. Fig. 186,

and d, illustrates the idea. At a is represented the would be obtained with a small current flowing in the conductor at fr, c, and d, the successive developments of a, &, c,

effect that

;

the magnetic field as the current increases. decreases in value, the reverse effect is true 170

When ;

that

the current is,

the mag-

INDUCTANCE AND COMMUTATION netic field will die 186. field

down, as illustrated by

d, c, 6,

171

and

Fig.

a,

This increasing and decreasing diameter of the magnetic about a conductor as the current increases or decreases

in value creates

an electromotive force in the conductor

and other conducting material in

close

proximity

X

C

Shows how magnetic

FIG. 186.

field builds

itself

to

the

"I"-- X

up about an

electric

conductor.

former.

In Chapter

was produced

force

magnetic

field,

X

we found out

that an electromotive

in a conductor

by moving it across a or when the magnetic field was moved so that

the conductor cut across the lines.

Consider two conductors placed close to each other, as in Fig. 187, 6, with a current flowing through one of them as

d FIG. 187.

Illustrates

mutual inductance.

indicated by the cross. As the current increases in value, the diameter of the magnetic field also increases, until the current

reaches a normal value, as illustrated by b, c, d, and e. The magnetic field of the live conductor L is expanding to the right by the dead conductor D. This is equivalent to moving conductor to the left across the magnetic field of conductor

D

L,

which gives the conditions for generating an electromotive namely a conductor cutting a magnetic field. The next thing to consider is the direction of the electro-

force

;

motive force developed in conductor D.

This

is

illustrated in

ELECTRICAL MACHINERY

172 Fig. 188.

It has already

conductor

L

D, which

is

been seen that the magnetic

direction, as indicated in Fig. 188

are in a

hand

field of

expands in a right-hand direction by conductor equivalent to moving the conductor in a left-hand

downward

;

also that the lines of force

Then by applying the

direction.

right-

rule for the direction of the electromotive force in con-

ductor D, Fig. 187, it will be seen that the fingers will point as indicated in Fig. 188, and the direction of the electromotive force produced in is opposite to that applied to L.

D

One thing

to

remember

is

that a voltage

is

produced in

D

only as long as the current is changing in value in L; as soon as the current reaches a constant value the lines of force also reach a constant value, cease expanding and are therefore no longer cut by conductor Z>, although conductor D is in the magnetic field when the current has reached a constant value, is no relative motion of the former across the latter. There is, however, a motion of the lines of force downward by conductor D, but this is only equivalent to moving the conductor upward parallel with the lines of force and therefore does not cut the magnetic field, but simply moves in it; con-

there

sequently no voltage

is

generated.

Voltage Induced by Decreasing Current. If the current in L is caused to decrease, then the circles of the magnetic decrease in diameter, and this will produce a leftdirection of the magnetic line in reference to conductor That is, if e is the original condition and the current de-

field will

hand D.

creases to zero,

a,

Fig. 187, will

show the condition when the

From

e to a the lines of force have been contracting in a left-hand direction past conductor D. This would be equivalent to moving the conductor in a right-

current has ceased to flow.

direction across the magnetic field. The equivalent conis shown in Fig. 189, and by applying the rule for determining the direction of the voltage when the direction of

hand

dition

the lines of force and equivalent direction of the conductor when the current in L are known, the voltage produced in

D

is

decreasing, will be seen to be in the

in L, as indicated in Fig. 189,

same direction

as that

INDUCTANCE AND COMMUTATION

173

Mutual Inductance. What we have just seen is that when the current is increasing in conductor L, it produces a voltage in conductor D opposite to that in L, and when the decreasing in L, it produces a voltage in D having the same direction as that in L. This voltage is produced by what is called "mutual inductance," and the voltage is said to

current

is

D

' '

' '

This is the princimutual inductance. ple of the induction coil and alternating-current transformer. What has been shown is that two coils with no electrical connection between them may have a voltage and current set up be induced in

in one

by

by passing a current that

the other. in value

When

direct current

is

changing in value through

used, the current the circuit.

is

by breaking and making

Direction

FIGS. 188 to 189.

is

varied

of [.Mf

Eule for determining direction of induced voltage.

Self -Inductance.

Another thing

to consider is the effect

of the magnetic field on the conductor itself. It has already been pointed out that the lines of force see'm to emanate from

the center of the conductor true,

it is

and expand outward.

If this

is

at once evident that the conductor cuts the lines of

force as they expand from the center to the outside of the wire. On the right-hand side of conductor L, Fig. 187, & to e, the

expanding in a right-hand direction, which would be equivalent to moving the conductor to the left. In

lines of force are

other words, conductor L is cutting the lines of force in the direction, just as a second conductor in close proximity

same

Since the foregoing is true, the wire carrying the current will have an electromotive force induced in it in the same direction as a second conductor in to L, as already explained.

174

ELECTRICAL MACHINERY

proximity to the first, and from what has been shown in Figs. 188 and 189 the direction of this voltage is opposite to the applied electromotive force when the current is increasing in value and in the same direction as the applied electromotive

close

when the current is decreasing in value. Or in other words, this voltage induced in the conductor carrying the current opposes any change in the value of the current. The

force

voltage induced in the conductor is said to be produced by self-inductance and, like mutual inductance, can take place only when the magnetic field is changing in value. The effects '

'

' '

of mutual and self-inductance enter into electrical problems under a great many conditions, therefore it is important that the principle explained in the foregoing be firmly fixed in the mind. One example of self-inductance is found in the coils

under commutation oh the armature of a direct-current machine, as will be seen from the following: Neutral Point on Commutator. In Fig. 190 the lines of force pass from the N pole into the armature core between points B and D; likewise under the S pole the magnetic flux passes from the armature core into the S pole between points A and C. It is only between these points that the armature conductors will be cutting the lines of force and therefore generating voltage, when the armature is revolved. Between points A and B on the left-hand side of the armature and C and D on the right-hand side, the conductors are outside of the magnetic field and are not cutting the latter therefore do not produce any voltage. The space between the polepieces where the conductors do not cut any line of force is called the neutral ;

It is always at this point that the point or neutral zone. brushes must be located on the commutator, because if they

are very far off the neutral, serious sparking will result. In general it is possible to shift the brushes slightly ahead of the neutral, that is, in the direction that the armature is turning, or to shift them slightly back of the neutral, against the direction of rotation, without seriously interfering with

the operation of the machine. In some cases the brushes can be shifted slightly ahead or back of the neutral with beneficial

INDUCTANCE AND COMMUTATION effect

upon

175

the operation of the machine, as will be explained

later in this chapter.

have already seen in Chapter X how a ring divided two parts acts to cause an alternating current generated

We into

in a coil revolving between the poles of a magnet to flow in one direction in the external circuit. It will be recalled that,

although the current was caused to flow in one direction, the current was of a pulsating nature; that is, flowed in waves.

FIG. 190.

Where

a

number

Ring-armature type generator.

of coils are connected to a commutator, as in

Pig. 190, the current not only flows in one direction in the external circuit, but also is maintained at a constant value.

This will be seen by considering what takes place at the brushes as the armature revolves.

How

Voltage in Armature

Is Maintained Constant. In shown after it has been revolved one segment from the position shown in Fig. 190. In Fig. 191 coil k, which was under the N pole in Fig. 190, has moved out from under the pole and into the neutral zone, while coil j, which is

Fig. 191 the armature

is

ELECTRICAL MACHINERY

176

in the neutral zone in Fig. 190, has

thus maintaining the same

number

moved

in

under the

N pole,

of active conductors

under

The same thing has happened under the S pole, this pole. where coil g, which is under this pole in Fig. 190, has moved out into the neutral zone and coil h has come in under the

maintaining the number of active conductors under

pole, thus

this pole constant

and consequently maintaining the voltage which in turn will cause a current of

at the brushes constant,

FIG. 191.

Same

as Fig. 190; armature turned one commutator segment.

constant value to flow in an external circuit

L

of constant

the process that is going on all the time in the armature as long as it is revolved. As fast as one armaresistance.

This

is

ture coil moves out from under a polepiece, another moves in to take its place, thus maintaining the number of active coils on the armature constant.

The process that takes place around the

coils

under com-

is, the coils in the neutral zone, is one of the most complicated operations in the machine. In Fig. 190 the current in coil I is flowing up through the plane of the paper

mutation, that

INDUCTANCE AND COMMUTATION and is

to the positive brush.

flowing from segment

At

177

the negative brush the current and down through the

c to coil

m

plane of the paper. In Fig. 191 the current in coil Z is flowing down through the plane of the paper and to segment e and then to the positive brush, and at the negative brush the curflowing in through segment ~b and to coil m, up through the plane of the paper. From this it is seen that the direction

rent

is

of the current in coils

FIG. 192.

Same

and

I

as Fig. 190

;

m

is

armature

reversed in Fig. 191 from

coils short-circuited

by brushes.

that of Fig. 190. In other words, when the armature revolves through an arc equal to the width of a commutator segment, that

is,

causes one segment to move out from under a brush to move in, the current in a coil connected to the

and another

segments that the brushes rests on is reversed. In changing from the condition in Fig. 190 to that in Fig. were short191, there was a period when coils I and circuited this is shown in Fig. 192. In this case the positive

m

;

brush rests on segments d and e and the negative brush on segments b and c. When the brushes are in this position, as

ELECTRICAL MACHINERY

178

far as the circuits in the armature are concerned there need not be any current flowing in coils I and m, since as shown in the figure, the current to the positive brush can flow directly from coils h and k without flowing through coil I. Likewise at is from the brush to coils g without passing through coil m. In other words, coils m are shuted out of circuit, until the brushes move onto

the negative brush, the current

and and

j I

segments & and e, as in Fig 191, where the current must flow in an opposite direction, in coils I and m, to that in Fig. 190. The foregoing might be easily accomplished if it were not for the property of self-induction, which is present in every electrical circuit when the current is changing in value. It was shown in the foregoing that when the current is increasing

in value in a conductor, the conductors cut the line of force set up by the current and induce a voltage that tends to prevent

the current from increasing in value, and when the current is decreasing in value, the conductors cut the line of force in a direction which creates a voltage that tends to keep the current flowing in the circuit. In other words, the effect of in-

duction

is

to

oppose any change in the value of the current in

the circuit. Effects of Self-induction on Commutation.

Let us see

what the result of self-induction is upon the armature coils under commutation, such as coils I and m in the figures. Start with Fig. 190 and consider only the positive brush. The brush moves off segment d onto segment e bridging across the insulation between the two segments, as in Fig. 192. If it were not for the induction of the coil, there would be no reason for the current flowing in coil I, Fig. 192, but when the circuited and the current starts to decrease, it

coil is shortis

prevented short must confor a induction and period by tinue to flow through the coil into segment d. If this continues

from doing

so

until segment d moves out from under the brush and segment e moves in, as in Fig. 191, then the current must not only cease flowing from coil I to segment d, but also reverse its direction

and build up to full value in the opposite direction, as in Fig. 191. In the latter case induction again tends to prevent the

INDUCTANCE AND COMMUTATION

179

current from building up in the opposite direction. The result of this is, if some means, which will be considered later, is not employed to make the current reverse in the coil under

commutation in the time required for the brush to pass from one segment to another, severe sparking at the brushes will take place. This is caused by the current not being able to reverse in the coil, in the time that the brush passes from one segment to another, and follows the brush across the insulation between the segments, similar to the way the spark is produced in a make-and-break ignition system on a gas engine. Another thing is that as segment d moves out from under the brush, the contact between the brush is getting smaller all the time until the brush leaves the segment. If considerable current is kept flowing from the coil under commutation into the segment that the brush

may

is leaving, such as coil I to segment eZ, Fig. 192, it increase the temperature of the trailing corner of the

brush to the point where it will glow. Commutation Period Short. The time during which the current must decrease from full value to zero and build up to full value in the opposite direction is very small. For ex-

ample, assume that the armature in the figure is revolving at 1,500 r.p.m., which is not an excessive speed. Since there are

=

24 segments in the commutator, 24 X 1,500 36,000 segments pass each brush per minute, or 600 segments per second. We have just seen that each time a segment passes a brush, the current reverses in a coil. Therefore, when a two-pole armature having 24 segments revolves at 1,500 r.p.m., the current must reverse in the coil under commutation in 1 / 600 part of a second. From this it is evident that it may be a somewhat

proposition to make the current properly reverse in the coil in such a short period. At the positive brush the current is flowing up through the difficult

coil

under commutation and must be reversed and caused

to

flow down, each time that a segment moves out from under the in. What would help to reverse would be an electromotive force induced in the

brush and another one moves the current coil

opposite to the direction that the current

is

flowing in the

ELECTRICAL MACHINERY

180 coil

under commutation

;

that

is,

if

the current

is

up through

the plane of the paper, the voltage will have to be downward to assist in changing the direction of the current. In the fig-

ure

all

S pole have an electromotive them down through the plane of the paper.

the conductors under the

force induced in

Therefore,

if

the brushes are shifted so that the positive brush tip of the S pole, as in Fig. 193, the coil

come under the under commutation will

FIG. 193.

Same

will

have a voltage induced in

it

that

is

as Fig. 190; brushes shifted to improve commutation.

When the opposite to the flow of the current in the coil. brushes are properly located, they will be in a -position where the Joltage generated in the coil will be just sufficient to reverse the Barren t

durmg he

period of commutation.

The foregoing is one way to obtain sparkless commutation and was the one usually relied upon in the early type of machines, but by improvement in design it has become possible to build generators and motors that will operate over their entire range from no load to full load with the brushes located exactly

between the polepieces without sparking.

It will be noted

INDUCTANCE AND COMMUTATION

181

that the brushes are shifted in the direction in which the

armature

is

revolving.

However,

this is true only of a gen-

erator.

When we consider the electric motor, it will be found that the brushes must be shifted against the direction of rotation, to assist in reversing the current in the coil under commutation.

CHAPTER

XIII

TYPES OF DIRECT-CURRENT GENERATORS Exciting Field Coils of Direct-Current Generators. In previous chapters we considered that the field coils of the generator were excited from an outside source that is, as in Fig.

.

-

;

FIG. 194.

.

Separate-excited shunt generator.

194, the field coils are assumed to be connected to some source of electric current, for exciting them, separate from the arma-

In alternating-current generators the field coils are always excited from a separate source of direct current, but in direct-current generators the field coils are in almost all cases

ture.

from the armature. There are two ways of one by connecting the field coils directly across the

excited directly

doing

this,

182

TYPES OF DIRECT-CURRENT GENERATORS brushes

that

is,

the field winding

is

in parallel with the arma-

and another by connecting the

ture, as in Fig. 195

183

field coils

in series with the armature, as in Fig. 196.

When

the field coils and armature are connected in parmachine is known as a shunt-con-

as in Fig. 195, the

allel,

nected generator; when the field coils and armature of a generator are connected in series, as in Fig. 195, it is known as a

machine. The shunt-type machine, or modifications of which will be considered later, is the type that is generally used, the straight series type being seldom used and then in only special cases. To simplify the connection in Fig. 195 and subsequent figures, the commutator will be shown on the out-

series it,

side of the

winding and the yoke will be dispensed with. With the machine that excites

Residual Magnetism.

its

question that arises is, How does the machine start to generate? If the machine were new and

own

field coils,

the

first

never had been in service

an

electric current

before', it

would not generate until

had been caused to flow through the

field

coils to magnetize the polepieces. When the field poles have been magnetized, they will retain a small percentage of the magnetism after the current has ceased to flow through the field coils. This generally amounts to about 5 per cent of the normal field magnetism. The magnetic flux which remains in

the field poles after the current has ceased to flow in the coils is called the residual magnetism. This residual magnetism is sufficient in a 110-volt

machine

to cause a*bout 5 or 6 volts to

be generated in the armature when running at normal speed and with the field coils disconnected from the armature, as in Fig. 197

;

in a 220-volt machine, approximately 10 or 12 volts due to the residual magnetism.

will be generated

How

Voltage

Is Built

up

in the Armature.

If the field

are connected across the armature, as in Fig. 195, and the latter revolves in the direction of the curved arrow, a small coils

voltage will be, as pointed out in the foregoing, generated in the armature windings. This small voltage, say 5 volts, will

cause a small current to flow through the field windings if in it will cause the field strength to be in;

the proper direction,

ELECTRICAL MACHINERY

184

creased above that of the residual magnetism and result in an in voltage.

In Fig. 195 the polarity of the residual magnetism

is

denoted by N. and 8, which will, for the direction that the armature is turning in, cause the right-hand brush to have This in positive and the left-hand brush negative polarity. turn will cause a current to flow through the field coils in the direction indicated by the arrowheads. By applying the rule for the polarity of a coil of wire with an electric current flow-

FIG. 195.

Shunt-type generator.

FIG. 196.

Series-type generator.

ing through it, it will be found that the field coils will have a and 8', which will be seen to be the polarity as indicated by

N

same

as the residual

r

magnetism in the

quently the current flowing in the

polepieces.

Conse-

field coils will assist in

mag-

netizing the polepieces, and the small current set up in the field coils by the 5 volts, which we assumed was generated due to the residual

magnetism in the

polepieces, will increase the number of lines

strength ; that is, there will be a greater of force entering and leaving the armature.

field

The armature

TYPES OF DIRECT-CURRENT GENERATORS

185

will therefore be cutting a greater number of magnetic lines, hence causing the voltage to increase, which in turn will cause the field current to increase, thus bringing about another increase in the field flux and also the voltage in the armature. This process continues until the machine is generating full

voltage.

FIG. 198

FIG. 197

Fie. 197.

FIG. 198.

Same as Fig. 195 Same as Fig. 195

j

;

field

field

winding disconnected. connections reversed.

Lines of Force in Field Poles Does Not Increase in Proportion to the Current Flowing in the Field Coils. The next question that naturally arises is why this process does not keep on indefinitely and the voltage continue to increase in value.

The answer

to this

is

found in the fact that the

lines of force

in the field poles do not increase in proportion to the current flowing through the coils. If

we were

to take a generator with the iron in the

magno residual magnetism in and connect a voltmeter across the armature terminal and

netic circuit absolutely dead, that it,

is,

ELECTRICAL MACHINERY

186

drive the machine at normal speed, it would be found that the voltmeter would give no reading, indicating that no voltage was being generated. However, if the field coils are connected to a separate source of voltage

and a small current caused

to

flow through the field coils, say 0.2 ampere, we would find that the generator would produce an electromotive force of, say 90

Then if we were to increase the current to 0.4 ampere, would be found that the voltage may not increase as much for the second 0.2 ampere as it did for the first. This, however, will depend somewhat upon the normal voltage of the machine. In this case assume the normal voltage to be 110 and that when 0.4 ampere was flowing in the field coils, the machine generated 115 volts. The foregoing is indicated on the curve Fig. 199. Here it is shown that if the field current is increased to 0.6 ampere, volts. it

04

Amperes FIG. 199.

in

05 06 Reid Coils

Direct-current generator voltage curve.

the volts will only increase to about 120, and beyond this point if the current is increased to 1.2 amperes, the voltage only increases to 124. For the first 0.6 ampere supplied to the field coils the

voltage increases from

to 120, but for the next 0.6

TYPES OF DIRECT-CURRENT GENERATORS ampere the

187

from 120 to 124, or an inThe foregoing indicates that the lines of

e.m.f. only increases

crease of 4 volts.

force in the polepieces increase rapidly at

current in the

field

coils

increases,

first,

a condition

but as the is

reached

Deyond which increasing the current in the latter will not This condition is cause any increase in the lines of force. called the point of saturation; that is, the iron is saturated with magnetic flux, just the same as a sponge becomes sat-

urated with water. Field-Coil Connections

and Armature Rotation.

relation exists between the connection of the field

the armature and the direction of rotation.

A

fixed

winding

to

It has already

been shown that the field-coil connections to the armature in Fig. 195 are such that the current flows through them from the armature, in a direction to make the field coils the same polarity as the residual magnetism in the polepieces, thus causing the machine to build up to normal voltage. However,

suppose we interchange the field-coil connections as in Fig. 198. In this case the polarity of the field coils, as indicated by N' and 8' is opposite to that of the residual magnetism, indicated by N and 8. Consequently, instead of the small current caused to flow in the field coils by the voltage generated due to the residual flux, increasing the field strength, it has the opposite effect and the machine cannot build up its voltage.

At

that if the polarity of the the machine connected as in magnetism reversed, 198 could build up. Considering Fig. 200 will show that Fig. this is not true. Since the residual magnetism is reversed, as first

thought

residual

it

may appear

is

N and 8,

the voltage generator in the armature is reversed; consequently, the current in the field coils is also reversed, as indicated by the arrowheads. This again brings

indicated by

the polarity of the field coil, as shown by N' and 8', opposite to that of the residual flux, and the machine cannot build up.

Therefore there

is

it is

evident that for the direction of rotation shown

only one

way

that the field coil can be connected to

the armature and have the machine generate, and that Fig. 195.

is

as in

ELECTRICAL MACHINERY

188

Reversing Rotation of Armature. direction 'of rotation

is

If

the

armature's

reversed, as in Fig. 201, then the field-

connections to the armature have to be reversed before the machine can build up. Assume the same polarity for the residual magnetism as in Fig. 195 then, since the direction of coil

;

rotation

reversed in Fig. 201, the voltage generated in the armature winding will be reversed, as indicated by the arrowheads. This voltage will cause a small current to flow through is

FIG.

200

FIG. 201

Same as Fig. 198; residual magnetism reversed. Same as Fig. 195; with direction of armature reversed.

FIG. 200. FIG. 201.

the field coils in a direction as shown, which gives the coils a polarity N' and 8' which is opposite to that of the residual

and the generator cannot build up to normal voltage. To produce a condition where the machine can build up

flux,

it

its

will be necessary to interchange the field connection

voltage to the armature terminals, as in Fig. 202. This allows the small voltage due to the residual magnetism to set up in the field coils a current that will give them the same polarity as the residual flux, and the machine will build up normal voltage.

TYPES OF DIRECT-CURRENT GENERATORS

189

The foregoing is an important point to remember when putting into service a new machine or one that has been repaired. After the field poles have been excited by sending a current through the field coils from an outside source, to the residual magnetism, if the machine does not

establish

build

up,

then the shunt-field

coil

connection

should

be

reversed.

FIG.

202.

Shunt generator rotating opposite from that with

One way

of

field coils

in

Fig.

195,

connected correctly.

knowing when the

field coils are

connected in

the right relation to the armature is as follows: First bring the machine up to normal speed with the field coils discon-

nected from the armature and note the voltage generated, which should, as pointed out in the foregoing, be about 5 or 6 for a 110-yolt machine, 10 or 12 for a 220-volt machine, etc. After doing this connect the field coils to the armature, and if

the voltage due to the residual magnetism decreases, the

field-

ELECTRICAL MACHINERY

190

connections must be reversed for the machine to come up normal voltage. For further information on changing the field connections of generators see discussion on Figs. 219 and 220 at the end of this chapter. Series-connected Generator.^In the foregoing we saw how a shunt generator was capable of building up a voltage from the residual magnetism in the fieldpoles when the field coils are connected to the armature in the proper relation, as coil

to

FIG. 203.

Shunt-type generator.

in Fig. 203.

In this case the

FIG. 204.

Series-type generator.

field coils are

connected across

the armature, therefore the latter is capable of causing a current to flow through the former whether the armature is

supplying an external load or not. In Fig. 203 the field circuit is from the positive brush through the field coils back to the negative brush without passing through terminals and

M

N, which lead to the external load, hence it is evident that the field circuit of a shunt-type generator is independent of the load circuit.

In the series-connected generator, as in Fig. 204, it will be seen by following around from the brush marked plus through the field coils that in order for the circuit through the arma-

TYPES OF DIRECT-CURRENT GENERATORS

191

ture and field coils to be completed an external load L must be and N, as in Fig. 205. In connected between terminals

M

other words, the series generator cannot build unless it is connected to a load.

up

its

voltage

coils of the series generator are connected in with the load and armature, it is at once evident that the cross-section of the conductors in the field coils must be large

Since the field

series

enough to take care of the full-load current of the machine. The field coils on the shunt generator are connected across the armature and are wound with wire of a size that will make the such a proportion as to produce the flux in the poleminimum expenditure of energy. The power excite to the field coils is generally from about 1 to required coils of

pieces with the

3 per cent of the output of the machine.

Ampere-Turns on Field Poles. To generate a given voltage, the armature must revolve at a certain speed and the magnetic field of the polepieces must have a definite value. To set up the lines of force in the magnetic circuit, it is necessary that a required number of ampere-turns in the field coils, say 6,000, be supplied. Ampere-turns is the number of turns in a coil of wire multiplied by the current in amperes that flows through it when connected to an electric circuit; that is, if a coil contains 2,000 turns and when connected to a 110- volt circuit, 3 amperes flow through

it,

then the ampere-turns are

2,000X3 = 6,000. If we assume the machine

in Fig. 203 to require 6,000 ampere-turns to excite the field coils sufficiently for the armature to generate 110 volts, and further assume that the full-load current of the machine is 100

amperes and requires 3 amperes to excite the field coils, then the number of turns in the field coils will be ampere-turns divided by amperes, or 6,000-^3=2,000, and the resistance of the wire in the field coils is volts divided by current, or 110-^-3=37 ohms approximately. If the series machine, Fig. 205, is assumed to have the same capacity as the shunt machine, Fig. 203, and requires the same number of ampereturns to excite the field coils as the shunt machine to generate 110 volts, then the number of turns of wire required in the field coils, since the total current is flowing through the field winding, will be 6,000-i- 100 = 60 turns, Since the total current flows through the field or 30 turns on each coil. coils and load in series, the combined resistance of the field coils and load can only be volts-i- amperes, or in this case, 110-MOO = 1.1 ohms.

ELECTRICAL MACHINERY

192

Now

we

are to keep the amount of power expended in the field coils machine down to approximately that of the shunt machine or 3 per cent, then the resistance of the field coils can be only 3 per cent of the total resistance, or 1.1X0.03 = 0.033 ohm, or the resistance of the

on the

if

series

37 field coils

shunt machine

is

= 1.121

times that of the series machine.

0.033

FlQ. 205

FIG. 205. FIG. 206.

Series generator connected to a load. Shunt generator connected to a load.

The foregoing shows the most prominent structural difference between the shunt and series type of machines. The field coils on the shunt machine are wound with a large number of turns of small wire having a comparatively high resistance and are connected in parallel with the armature. The field coils of the series machine are wound with a small number of turns of large wire, consequently have a low resistance and are connected in series with the armature. However, the size of the conductors in either case varies with the size and the voltage of the machine.

The comparison

of the field coils in the foregoing is not first case we assume 110

absolutely correct, because, in the

TYPES OF DIRECT-CURRENT GENERATORS volts at the brushes

and

in the series

193

machine we have assumed

that the total pressure generated is 110 volts. However, the comparison is close enough for all practical purposes and eliminates a lot of calculation.

evident that with the series generator if the field to be maintained constant, consequently the voltstrength at the brushes at a constant value, the load also will have age It

is

is

maintained at a constant value. This is generally a thing to do, since the load on a generator is usually made up of a number of different devices used for different

to be

difficult

sizes and types, which are connected when wanted and disconnected when not

purposes and of different the

to

circuit

required. The devices also require approximately a constant voltage for their operation. Such a condition cannot be met very successfully by the series generator.

From what we have already seen of the evident that the load on the machine does not

Load on a Shunt Generator. shunt generator,

it

is

For example, in Fig. 206 is given a shunt affect the field circuit. generator supplying a load of four resistances, r\, r, r 3 and r 4 each of 4 ohms, in parallel. If we assume that the armature develops 100 volts and neglecting the effect of the armature resistance, the current i flowing in each section of the load

amperes

is

in the four circuits.

i=

E = 100 =

If

25 amperes, or a total of 100

one resistance

is

disconnected from the

the current supplied to the load will be 25 X 3 = 75 amperes, and only two are connected, the current delivered to the load by the ar-

circuit, if

mature is 50 amperes, and for one resistance, 25 amperes. Under any one of the conditions the current flowing in the field coils will remain practically constant since, as shown, this circuit is independent of the load. Consequently, the value of the field current is not affected by the load, except as the voltage is caused to vary slightly by the load current and This latter factor will be considered further resistance of the armature.

on

in this chapter.

Load on a Series Generator. Now consider what would happen if we varied the load on the series generator, Fig. 207, the way that it was changed on the shunt generator, Fig. 206. In Fig. 207, if we assume the machine to be generating 100 and that 100 amperes is flowing in the amperes is passing through the field coils.

volts

circuit,

then 100

If one section of

ELECTRICAL MACHINERY

194

the load was taken off

and

if

the voltage at the armature ter-

minals remained constant at 100 volts, as was assumed in the shunt machine, 25 amperes would flow through each of the three resistance elements, as in Fig. 206. But with the series

machine the pressure will decrease since the current has been decreased in the field coils, consequently the current will decrease in the different elements connected across the armature terminals.

FIG. 207. FIG. 208.

From

this it is evident that as the load is

Series generator connected to a multiple load.

Rheostat in

field circuit

of shunt generator.

decreased on the series machine the voltage is decreased and current through each individual load also decreases;

the

whereas, on the shunt generator the total load may be varied, but the current in the individual loads and the voltage

remains practically constant. What we have just seen has practically eliminated the series type of machine from commercial use in preference to the shunt type or modifications of this latter type. Voltage Control on Shunt Generator. Direct-current

generators are generally designed so that,

if

the shunt-field

TYPES OF DIRECT-CURRENT GENERATORS

195

is connected directly across the armature, as shown in Fig. 203, they will at rated speed develop about 120 per cent normal volts, that is, a 110-volt machine will generate

winding

about 125 or 130 volts. The voltage is then adjusted to normal by connecting an adjustable resistance in series with the The current through the field field circuit, as in Fig. 208. coils is adjusted by means of this resistance so as to produce normal volts. Then any slight variation in the voltage, due to changes in load or otherwise, can be taken care of by varyThe resistance coning the resistance in the field circuit.

nected in series with the

field coils is called

of Direct-Current Generators.

Diagrams diagram of a shunt generator or motor.

a

field rheostat.

In Fig. 209

In this

all

is

that

a is

the circuits; the armature is indicated as a segmental ring and the field winding as a spiral. However, it will be seen that the field circuit in Fig. 209 is in parallel with the

indicated

is

armature, as in Fig. 203. In Fig. 210 the field winding is connected in series with the armature as in Fig. 204, making one

through the machine in either case. The diagrams, and 210, provide a convenient means of representing the circuit through electrical machinery, and will be used circuit

Figs. 209

many

times in future chapters.

Effects of Loading a Shunt Generator on Voltage

Regulation.

In

Fig. 211 the field coils are shown excited from a source separate from the armature, so that any variation in the voltage at the armature terminals will not affect the strength of the magnetic "field. Assume that the

ohm resistance and generates 115 volts on open circuits, as in Fig. 209. Now, if a resistance of R' = 5 ohms is connected across the terminals of the generator, as in Fig. 211, the total resistance of the circuit will be R equals that of the armature and external circuit in

armature has 0.23

series,

or

# = r+#' = 0.23+5 = 5.23

flow in the circuit

is

/=f

.

As has been explained

=

~

ohms, and the current that

= 22

will

amperes.

in Chapter VI, part of the voltage produced be used up in the armature winding to cause the This voltage e is equal current to flow through this section of the circuit. to the resistance of the armature times the current; that is, e=r/=0.23 X 22 = 5. 06 volts. From this we see that when 22 amperes is flor ing

in the

armature

will

ELECTRICAL MACHINERY

196

in the circuit, there is 5.06 volts

drop in the armature winding. Hence e = 115 5.06 is E a =E

the available voltage at the armature terminals = 109.94 volts, as shown.

Consider what would be the

connecting a second resistance

effect of

The across the generator terminals, as shown in Fig. 212. = 5-f-2 joint resistance of r' and r" is R' equals one-half that of r', or #' = 2.5 ohms. Then the total resistance of the circuit is the joint resistance

of 5

ohms

of the external circuit

and that

of the

armature winding, from which

R = R'+ r = 2.5X0.23 = 2.73

ohms,

and the current ET

/= To e

-I

=

R

r

-j

2.73

= 42 amperes

approximately.

cause the current to flow through the armature will require a voltage This will leave a voltage of E a =E-e = U5

= rl= 0.23X42 = 9.66 volts. 9.66 = 105.34 volts available

From what we have

at the armature terminals, as indicated.

.seen in Figs.

211 and 212,

it is

evident

increased on a shunt generator the voltage at the armature terminals decreases. The voltage generated by the armature would also, to a certain extent, decrease if that as the load

is

the field coils are connected to the brushes, as shown in Fig. 209. For the reason that as the voltage decreases across the

armature terminals the current will be decreased in the

field

coils, consequently the number of lines of force will be reduced. In Fig. 211 with a resistance of 5 ohms connected

across the armature 22 amperes flowed through the circuit, kvnile in Fig. 212, where two resistances of the same value are

connected in parallel, only 21 amperes is sent through each resistance, showing that as the load increases on a shunt gen-

some means

taken to maintain the voltage constant, the current will decrease in each circuit as more load is connected to the generator. How Voltage May Be Maintained Constant. One way of erator, unless

is

maintaining the voltage practically constant would be to design the generator for about 20 per cent over voltage and connect a rheostat in series with the

field

winding, as in Fig.

213, to reduce the field current to a value where normal voltage would be generated at no load then, as the voltage falls off because of an increase in load, sections of the rheostat can ;

TYPES OF DIRECT-CURRENT GENERATORS

197

ELECTRICAL MACHINERY

198

be cut out of circuit so that the

field

current will increase to a

value that will cause the generator to produce sufficient pressure to maintain the voltage at the armature terminals constant. rheostat cut out of circuit, as in Fig. 214, 135 volts, and with part of the rheostat cut in series with the field windings, as shown in Fig. 213, the

For example, with the assume that the machine

field

will generate

voltage decreases to 115. Then, neglecting the effect of the decrease in voltage at the armature terminals, due to increase of load, on the field winding and connecting a 5-ohm resistance across the armature terminals, as in Fig. 215, the current in the circuit will be approximately 22 amperes and the voltage will drop to 109.94, as in Fig. 211. Now to bring the voltjge back to normal, some of the field rheostat can be cut

out as in Fig. 216. This will increase the current in the field coils and in turn increase the field strength, so that the armature conductors will be cutting a greater number of lines of force and producing a great voltage; as is shown in the figure, the voltage has been increased to normal, or 115.

With 115

volts available at the armature terminals, the current in the 7TT

external circuit

and the

is

/=

R

-I

-I

pr

= 23

= 5

amperes, instead of 22, as in Fig. 215,

volts drop in the armature

is

e=rl= 0.23X23 = 5.29

volts.

Therefore, for the armature to maintain 115 volts at its terminals with a 23-ampere load, it will not only have to generate the 115 volts availits terminals, but also 5.29 volts to cause the current to flow a +e = 115+5.29 through the resistance of the windings or a total of

able at

E=E

= 120.29

volts.

After the voltage had been adjusted to 115 at the armature terminals with a load of 23 amperes, if the load was taken off and the field rheostat not changed, the voltage would increase to 120.29 volts, or the total of that generated in the armature. Although, in Fig. 216, only 115 volts is available at the armature terminals, nevertheless, the machine is generating 120.29 volts; 5.29 volts is used up in the armature winding.

As soon as the load is taken off, there is no current flowing through the winding to use up the 5.29 volts and it becomes available at the brushes. To bring the volts back to normal again it will be necessary to cut the resistance back into the field circuit, as in Fig. 215.

From what we have shunt generator

seen

it is

evident that

if

the load on a

varying, the voltage will fluctuate accordOf course, these fluctuations, if they do not

is

ing to the load. occur too rapidly, can be taken care of by the operator adjusting the field rheostat. However, a better way of doing this, if

TYPES OF DIRECT-CURRENT GENERATORS

199

would be to incorporate some automatic means in the construction of the machine to maintain the voltage constant.

possible,

Compound- Wound found out that

if

Generator.

the load

is

In

the

foregoing

we

increased on a series-connected

generator the voltage will increase, and decrease as the load decreases. Taking advantage of this fact provides a means of obtaining a close voltage regulation

on direct-current gen-

done by constructing what may be called a combination of a shunt and series machine, or, as it is known,

erators.

a

This

is

compound-wound generator.

This connection

is

shown

in

From this figure it will be seen that one field connected in series with the armature, as in Fig. 204, and a second field winding connected across the armature, as wind-

Fig. 217.

ing

is

The shunt-field winding provides the flux to generate about 110 to 115 per cent normal voltage, the 10 or 15 per cent excess volts being taken by the field rheostat. The in Fig. 203.

winding sets up the flux necessary to generate the additional voltage to compensate for the volts drop through the armature due to the load current and the resistance of the

series-field

winding. In Fig. 217, if the armature is revolved in the direction of the curved arrow, a voltage will be generated in the winding of a polarity as indicated and a current will flow through the shunt-field windings in the direction shown by the arrowheads.

This voltage can be regulated to normal by adjusting the field we will assume, to 115 volts. With no load on

rheostat, or as

the machine no current is flowing through the series-field winding, although some machines are connected so that the shunt-field current flows through the series-field winding. If a resistance is connected across terminals and N, as

M

in Fig. 218, of such value as will allow a current of, say, 25 amperes to flow, as indicated, this current passes through the series-field winding and will increase the number of lines of

force entering

and leaving the armature, consequently the

voltage generated in the armature conductors will be increased. On the other hand, the current flowing through the

armature will cause a certain voltage drop in the winding.

ELECTRICAL MACHINERY

200

Now

if

5 volts

is

required to cause the current to flow through

the armature winding, and the series-field amperes-turns cause the magnetic field to increase in value to where the armature will generate 120 volts, then the 5 additional volts will just compensate for the loss in the armature and the volts at the

armature terminals will be maintained constant.

If the cur-

rent supplied to the load is increased to 50 amperes, then the current through the series-field winding will increase to 50

amperes, which in turn will increase the number of lines of

Fm.

Via. 217

FiGS. 217 and 218.

21 o

Compound-wound generators.

and leaving the armature and again cause the generated to build up and compensate for the drop in

force entering volts

the armature, thus

keeping constant e.m.f. at the brushes.

The foregoing characteristic of the compound generator, which is nothing more nor less than a shunt generator, having addition to the shunt winding, a series-field winding on its polepieces, to automatically maintain the voltage approxiin

mately constant at its terminals, has caused this type of machine, with certain modifications, to be adopted almost universally for generating direct current. Due to the iron in the polepieces

becoming saturated, the

lines

of

force

do not

TYPES OF DIRECT-CURRENT GENERATORS

201

increase in proportion to the ampere-turns on the field coils, thus making it impossible to design a compound generator that will maintain absolutely constant voltage from no load to full

This subject will be discussed in the next chapter.

load.

Crossing Field Connections of Compound- Wound Generator. In the foregoing it was shown that when the field coils of a generator are connected so that their magnetomotive force opposes that of the residual

magnetism in the polepieces, The statement was

the machine cannot build

up

made

this condition is to "cross the field

that the

remedy for

TIG. 219.

connections.

done, is

and

"

its

voltage.

Large compound-wound generator

In a shunt machine this can generally be easily winding in a compound machine

as far as the shunt

concerned the foregoing

is

always true.

But

in the com-

generator, to keep the series winding the same polarity as the shunt, the leads of the former must also be crossed, if

pound

However, if the machine is of 100-kw. capacity or above, it not infrequently happens that the series-field leads are heavy copper bars brought to terminals, as shown in Fig. 219, which makes it impossible to cross these leads.

this is possible.

If

it

machine

is

necessary to cross the field connections for the shunt connections can be crossed and

to build up, the

ELECTRICAL MACHINERY

202

the generator will come up to voltage, but if the field windings are in opposition when the load is thrown on, the voltage will

decrease very rapidly as the load increases, owing to the effects of the series-field winding.

demagnetizing It

may

be possible to cross the armature connection instead If this can be done, the machine will come up to

of the field.

voltage and the field windings will have the correct polarity. But at best this generally results in a bad arrangement of the leads around the brush gear, and in many cases the armature leads cannot be crossed.

FIG. 220.

The

Same

as Fig. 219, with brushes shifted 60 degrees.

best solution of the

problem

is

obtained by shifting the

brushes from one neutral position to the next, which means that the brushes will be shifted 90 deg. on a four-pole machine,

60 deg. on a six-pole machine, etc. and 220 will make clear this idea.

A

comparison of Figs. 219

The brushes in Fig. 220 have been shifted clockwise around the commutator 60 deg. from that in Fig. 219. This will reverse the current through the field windings of Fig. 220 and is the equivalent of crossing the leads of both field windings in Fig. 219 and it also simplifies

the problem.

CHAPTER XIV DIRECT-CURRENT GENERATOR CHARACTERISTICS Voltage Curves of Shunt Generator. Electric generators and motors act in certain ways unde~r given conditions for ;

example, as the current

is

increased in the field coil of a shunt

150

o EiOO

O

O

75

!

o

Q2.

FIG. 221.

0.4

Amperes m

06

0.8

12

,

Field Coils

Direct-current generator voltage curve.

generator, the voltage will increase in value until the iron in the polepieces becomes saturated. Beyond the saturation point the voltage remains practically constant irrespective of

the value of the current in the field

coils.

field coils,

By

plotting the

armature terminals against the amperes in the as explained in Chapter XIII the result will be a

volts at the

curve similar to that shown in Fig. 221. 203

ELECTRICAL MACHINERY

204

Similarly, the effect of the load on the voltage of a generator may be shown in a curve. This is done by connecting a

voltmeter across the armature terminals and an ammeter in

shown in Fig. 223 and after adjusting the voltage to normal, say 110 volts, put a load on the machine of, assume, 30 amperes, as indicated in Fig. 224. This, as shown in Chapter XIII, will cause the voltage to drop at the series in the circuit, as

armature terminals:

;

owing

First,

to the

armature, and second, the current in the

resistance of the

field coils will also

120

no 100

f E60

|

60

I 60

40

20

60

60

100

120

140

160

160

eOO

ttQ

40

Amperes Load-voltage curve of shunt generator.

FIG. 222.

be slightly reduced because of the decrease in volts at the armature terminals. Assume that the volts at the armature terminals decrease

1.5, this will

and

leave

Ea =

108.5 volts avail-

on curve A, Fig. 222, which is obtained by taking the 30-ampere division at the base of the curve and running up vertically until it intersects the horizontal line running out from the 108.5-volt division, as able at the load

will give point a

indicated by the dotted lines. If the load is now increased to 60 amperes, the volts at the armature terminals will further decrease, say to 107; then

DIRECT-CURRENT GENERATOR CHARACTERISTICS

205

plotting the load current of 60 amperes against the voltage at the armature terminals, 107 volts, gives point b on the curve. Increasing the load to 90 amperes will cause the voltage to drop accordingly, or, as shown at point c on the curve, to be 105.5.

Now,

if

the load

decrease in voltage

is

is

further increased, a corresponding However, it is evident that

obtained.

cannot keep on indefinitely, because if it did, eventually a point would be reached where an infinitely large current would be obtained on an infinitely small voltage.

this process

Fia. 225

FIGS.

223 to 227.

FIG. 226

Fia. 227

Diagrams of shunt-connected and series-connected generators.

What

actually happens in a shunt generator is that the decreases as the current is increased up to a certain voltage value and then both volts and current decrease to zero or

approximately so. This is shown on the curve when the current has increased to 241 amperes, the volts have dropped to about 55. At this point, if the resistance of the circuit is ;

further decreased to increase the current, the voltage and current begin to decrease and come back to zero, or theoretically so.

However, on account of the residual magnetism in the

polepieces maintaining a small voltage at the armature terminals, the volts and current will only approximate zero.

ELECTRICAL MACHINERY

206

What has just been stated regarding the shunt machine indicates that if it was short-circuited, the voltage and current would drop

to zero

and no harm would be done.

This

is

true

of the self -excited shunt generator, but not of the other types,

as will be seen in the following Volts Drop in Armature With Fields Separately Excited. If the field coils of the shunt machine are energized from :

an outside source,

as in Fig. 225, then the field current will be

maintained constant irrespective of the load. Consequently, the voltage generated in the armature will remain practically constant, and the volts drop at the armature terminals will be

armature resistance only. Therefore, the volts at the armature terminals will not decrease so rapidly as when

due

to the

the field coils are connected in parallel with the armature. The resultant curve for a separate-excited shunt generator will be similar to the curve B, Fig. 222.

Voltage Curve of Series Generator. In the seriesconnected generator, Fig. 226, the machine cannot produce

any voltage when it is disconnected from the load except that generated due to the residual magnetism in the polepieces. Consequently, at no load the voltage of a series machine is approximately zero, as against the shunt machine in which the voltage is at a maximum value at no load. By connecting a voltmeter and an ammeter to the series machine, as in Fig. 227, and taking readings for different loads, a curve will be obtained as in Fig. 228. It will be seen that the shape of the curve from zero to point a approximates the shape of the saturation curve, Fig. 221, from zero to point a. The series generator as its voltage builds up with the load

has not only to produce pressure to cause the current to flow through the external circuit, but also through the armature

drop in the armature and field windings varies as the product of the current in amperes and the resistance of the windings in ohms. However, the total voltage generated in the armature winding does not increase

and

field

windings.

The

volts

Referring to Fig. 221, it a on the curve, the increase in

as the current in the field winding. will be seen that

up

to point

DIRECT-CURRENT GENERATOR CHARACTERISTICS volts

is

beyond

207

quite rapid as the field current is increased, this point the increase is

but

very slow, being practically

upper end of the curve. It is this latter fact that makes the voltage of the series generator decrease above a

zero at the

certain load.

In Fig. 228 the

20-ampere load energizes the field coils armature terthe load is increased to 40 amperes, the volts first

to the extent that 50 volts is generated at the

minal.

When 120,

110

100

o 90

r I

eo

* fi
ro

20

40

60

60

100

!20

140

160

180

200

220

240

260

Amperes PIG. 228.

Load-voltage curve of series generator.

only increase to about 73 and at 60 amperes about 91

volts,

until at 130 amperes, the maximum, or 114 volts, is developed at the armature terminals. From this it is seen that the volts

armature terminals increase rapidly at first, but that the increase becomes less for a given number of amperes increase until an increase in load does not cause the volts to

at the

become greater but

less,

as in this case,

when

the load

is

made

higher than 130 amperes. if we assume the resistance of the armature and field windings ohm, then with a 20-ampere load on the machine the volts drop equal amperesXohms, or 20X0.2=4 volts; that is, the armature is

Now,

to be 0.2 will

ELECTRICAL MACHINERY

208

actually generating 54 volts when supplying 20 amperes to the external circuit, but 4 volts is used up to cause the current to flow through the armature and field windings; therefore, only 50 volts is available at the

armature terminals. At a 60-ampere load the volts drop in the armature is 60X0.2 = 12 volts. Hence, the armature is generating a total voltage

12 = 103 volts. When the load has increased to 130 amperes, the volts drop in the armature is 130X0.2 = 26 volts, and there is generated 114+26 = 140 volts. However, 26 volts is used up in the machine's windings, consequently only 114 is available at the armaat this load of 91

ture terminals.

amperes.

+

Assume that the load

is

Then, the drop in the armature

increased from will

130 to 180 be 180X0.2 = 36 volts.

Further, assume that this increase in the load only causes the total voltage to increase to 143.5. Then, the available volts at the armature terminals is 143.536 = 107.5. Hence, it is seen that the increased volts generated in the

armature due to the increased load

is

not enough

to compensate for the increased drop in the windings, and the available volts decrease with an increase in load. Hence, it is evident that the volts at the armature terminals on- a series generator will increase in value with an increase in load until the iron in the magnetic circuit is near saturation; beyond this point the volts begin to decrease with an increase of load.

External Characteristic Curves. The curves, Figs. 222 228, are sometimes referred to as external characteristic curves of the generator, from the fact that they are plotted from conditions existing outside the machine. If the voltage

and

values existing in the armature windings were used in the curves, we would have to add the volts drop in the armature at the different loads to the voltage at the brushes corresponding

which would have given a higher pressure than that indicated on the curves in the figures. load-voltage curve plotted by using the total voltage generated in the to these loads,

A

armature instead of that at the brushes

is

called

an internal-

characteristic curve.

Voltage Curve of Compound Generator. With a series winding on the polepieces along with a shunt winding, as in the

compound

generators, Fig. 229, the load-voltage curve shunt char-

will, to a certain extent, be a combination of the

and the series curve, Fig. 228. In the compound-connected machine, Fig. 229, current is flowing through the shunt-field winding only, and this is adjusted by

acteristic curve, Fig. 222,

DIRECT-CURRENT GENERATOR CHARACTERISTICS

209

the field rheostat to give normal voltage at the armature terminals. When a load is connected to the machine, as in Fig. 230, the current passing through the armature will tend to cause the voltage at the brushes to decrease, but the load current flowing through the series winding will increase the

strength of the magnetic field and cause a greater voltage to be produced to compensate for the drop in the armature and

M FIG. 229.

Diagram of compound-wound generator.

series-field winding. If we assume that the machine is normally generating 110 volts at no load, the density of the magnetic circuit would correspond to point A on the mag-

netization curve, Fig. 231. When the machine is carrying full load, if the current flowing in the series-field winding inon the creases, the magnetic density to correspond to point

B

curve, then the armature will be generating about 124 volts; that is, the voltage generated in the armature has increased from 110 to 124, or 14 volts. Now, if the volts drop in the

ELECTRICAL MACHINERY

210

armature, from no load to full-load, is only 14, then the volts at the armature terminals at full load will be the same as at no load.

Series

Compensate for Volts Drop in Armathought it may seem an easy matter to pro-

Winding

At

ture.

first

to

portion the series-field winding so that for the drop in the armature circuit.

it

A

FIG. 230.

will just

compensate

further consideration

Diagram of compound-wound generator connected

load.

to

If of the curve, Fig. 231, will show that this is impossible. the volts indicates the no-load voltage and point generated in the armature at full load, point 0, half-way

point

A

B

A and B, will indicate the volts at half load. But A to C the voltage has increased from 110 to 120, whereas

between

from from

B

to

C

it

has only increased from 120 to 124, or 4

against 10 on the

first

half of the load.

The

volts

volts,

drop in the

armature is proportional to the amperes, consequently if load current causes a drop of 14 volts, then half full-load

full-

DIRECT-CURRENT GENERATOR CHARACTERISTICS

211

rent will cause 7 volts drop. But with half full-load current flowing in the series-field winding, in this case, it caused 10 volts increase, therefore, the voltage at the brushes is 3 volts

higher than at no load. What has really happened to the voltage of the compound generator from no load to full load is indicated in curve B, Fig. 232, assuming 200 amperes full load.

Here it is shown that although the voltage is the same at full load as at no load, nevertheless, it has not been constant between these points, increasing in value during the first half

es

o 100

7

QZ

04

0.6

Amperes FIG. 231.

of the load

in

0.6

L

field Coils

Direct-current generator voltage curve.

and then decreasing

to normal again during the not Furthermore, possible to design a compound generator that has a constant voltage from no load to full load. However, conditions similar to that indicated by last half.

it

is

curve B, Fig. 232, can be approximated. Voltage at Terminals of Compound Generator.

A

com-

pound generator that develops the same voltage at full load as at no load is said to be flat-compounded. It is possible, by proportioning the shunt- and series-field winding, to design a compound machine where the voltage will increase from no load to full load, as indicated by curve A, Fig. 232.

ELECTRICAL MACHINERY

212

For example we assume that the

full-load current causes 14 volts

drop

the current flowing through the series-field winding caused a 20-volt increase in the armature, then the voltage at the armature terminals will be 20 14 = 6 volts higher at full load than at

in the armature.

On

if

When

the voltage of a compound generator increases from no the machine is said to be over-compounded. the other hand, suppose that when full-load current is flowing

no load. load to

Now,

full load,

in the series-field winding

it

caused only 8 volts increase;

then, since

ICO

no 100

90 60 70

60 50

40 30 EO 10

20

FIG. 232.

40

60

60

100

120

WO

160

ISO

Compound -wound generator

200 220 240 260

load-voltage curves.

is 14 volts drop in the windings at full load and only 8 of these volts =6 are compensated for, the pressure will decrease at the brushes, volts. Such a condition is represented by curve (7, Fig. 232. A compound generator the voltage of which decreases from no load to full load

there

148

is

known

as being under-compounded.

Proper Amount of Compounding. Another feature in obtaining the proper amount of compounding of the generator This is the proper number of turns in the series winding. of a made is machine in the heavy copper large-sized winding bar, as in Fig. 233, therefore, the terminals will have to come out on opposite sides, so that the connection can be made conveniently between the coils. This means that the minimum number of turns in the coil must be 1.5. Then, to maintain the proper position of the coil 's terminals the number of turns will have to be 1.5, 2.5, 3.5, etc. One-half turn in the series-field

DIRECT-CURRENT GENERATOR CHARACTERISTICS winding at

first

importance.

thought

may

However, when

213

not seen to be of any serious considreed that in a 200-kw.

it is

machine the normal full-load current is approximately 2,000 amperes, and this current flowing through one110- volt

half turn gives 1,000 ampere-turns, the effect that only a small fraction of a turn in the series winding will have upon the

voltage of such a machine at once becomes apparent. Here again, in the design of the series winding,

it is

im-

by a coincidence, that the correct number of turns can be obtained. The result is that as the series-field

possible, except

winding machine

is

is

designed on most over-compounded.

compound

FIG. 234.

Large-capacity

the

LOAD

FIG. 234

FIG. 233

FIG. 233.

generators,

series-field

Compound generator showing

coil for

compound generator. compounding shunt.

location of

For example, assume that in working out the design of a it is found to give the proper amount of compound-

machine

winding. In turns and 3.5 only using or machine the using 4.5 slightly under-compounded, having turns and over-compounding the machine. The latter is the

ing, that 3.8 turns will be required in the series this the designer has the choice of

Due to imperfections in the materials and workmanship, the machine may vary somewhat from what was expected of it. In fact, it is practically impossible to build two machines after the same design and from the same lot of stock and have them both possess identically

best course for several reasons.

the same characteristics.

Therefore,

it is

good policy to use a

liberal design in the series-field winding, since there is a simple

ELECTRICAL MACHINERY

214

means for adjusting the ampere-turns of this winding when the machine is over-compounded. This consists of connecting what is known as a compounding shunt directly across the series-winding terminals, as shown in Fig. 234. Compounding Shunt. After the machine is built and in operation in the shop, a test is made to find out the amount ,of resistance that must be connected across the series-field terminals to give the required compounding, and then a shunt is made for this purpose and connected to the terminals of the series winding.

through the

Then, instead of

series

all

the load current passing

winding, only part of

it

does,

depending

upon the resistance of the shunt. If the full-load current of the machine is 1,000 amperes and only 800 amperes are required to compensate for the volts drop in the series and armature windings, at 'full load, then the shunt is made to have a resistance four times as great as that of the

series

winding, so that when it is connected in parallel with the series winding one part of the current will pass through the shunt and four parts through the series winding, or any degree of over-compounding may be obtained up to the maximum by increasing the resistance of the shunt. Direct-current generators have been built for railway work in which the voltage increased from 500 at no load to 550 at full load, this increase in voltage being used to compensate for the volts drop in the feeders.

Short-Shunt and Long-Shunt Connection. In passing, may be called to the way that the shunt-field windIn Figs. 229 and 230 the shunt winding are connected. ings attention

connected directly to the armature terminals. This is as a short-shunt connection. In Fig. 234 the shunt winding is connected directly across the series winding and is

known

armature in

series, so

that in this case the shunt-field current

This is known as a passes through the series winding also. long-shunt connection. Since the shunt-field current is only a very small percentage of the total load of the machine, it it makes little difference which connection

evident that

used, the choice being

anything

else.

is is

more a matter of convenience than

CHAPTER XV LOSSES IN DIRECT-CURRENT MACHINERY Loss in Transformation of Energy. Whenever energy is changed from one form to another, there is always a loss in For example, the amount of energy the transformation. transmitted by the steam to the cylinder of an engine in the form of heat is not all available at the flywheel to do useful work.

A

large percentage of the energy actually supplied to the engine is lost in the exhaust, in radiation from the surface of the cylinder and in overcoming the friction of the

moving

parts, etc.

What

has taken place in the engine is, the energy in the steam has been converted into a mechanical form of energy which may be used to do the mechanical work of driving any kind of machinery. If the engine is used to drive an electric generator, then we will have another transformation of energy that is, the mechanical energy transmitted to the ;

engine's shaft or flywheel will be converted into electrical energy and transmitted through the circuits to the devices

In this transformation from a supplied by the generator. mechanical to electrical energy there is also a loss just as in the steam engine; that

is,

if

the energy delivered to the

engine's flywheel capable of developing 100 hp., then less than 100 hp. will be delivered to the circuits. Part of the power developed at the engine shaft will be expended in overis

coming the

friction of the

citing the field

coils,

moving parts of the generator, exthe losses due to the resistance of the

armature circuits and eddy-current and hysteresis Friction Losses.

The

losses.

friction losses in a direct-current

machine consist of the friction of the bearings, brushes on the 215

ELECTRICAL MACHINERY

216

commutator and the friction of the air upon the revolving The last item is usually known as the windage losses. The total friction losses amount to about 6 per cent of the capacity of the machine for a 1-kw. unit to about 3 per cent for a 1,000-kw. unit. These may be considered as the mechanical losses of the machine that is, they represent mechanical power that has been supplied to the generator and that has not element.

;

been converted into electrical power, but has been expended in doing the mechanical work of overcomnig the friction of the generator.

Current to Excite the Field Coils. The current that is used to excite the field coils represents electrical power that has been generated in the armature, but is used up within the

machine to energize the for doing

work outside

not available

field coils, therefore is

of the machine.

The amount

of

power

required to energize the field coils of direct-current machines is about 6 per cent of the total output for machines of 1-kw. capacity to about 1.4 per cent in 1,000-kw. sizes.

Since the energy expended in the field rheostat is also charged up against field losses, the power loss in the shunt-

winding is practically constant, being only changed slightly by the hand adjustment of the rheostat. The losses in the shunt-field winding are therefore equal to the volts at the armature terminal times the current supplied to the field coils. The energy expended in the field coils is sometimes

field

2 referred to as the excitation losses or I R losses in the shunt-

field

winding; that

is,

the loss in the field coils

is

equal to the

square of the current times the resistance of the field coils that of the section of rheostat in series with the coils.

and

For example, the total resistance of a shunt-field circuit is 72 = 27.5 = ohms, and the voltage at the armature terminal is # 110; then the 1 1 O = 4 amperes, and the watts current flowing in the field coils is I = = |7T

R

W =#7 = 110X4 = 440. 27.5

= 440, which

27.5

are also

W=/ # = 4 2

2

X27.5 = 4X4X

gives the same result as the former method.

Copper Loss

we found out

The watts

in the

Armature.

In the previous chapters

that a part of the voltage generated in the

LOSSES IN DIRECT-CURRENT MACHINERY

217

armature was used up in overcoming the resistance of the armature windings to the flow of the current. This also represents a loss of power supplied by the prime mover to the This loss is usually called the armature copper generator.

PR

or loss, and is one of the chief factors in increasing the temperature of the machine. The power loss in the armature copper is equal to the voltage drop through the armature winding times the current supplied by the armature it is also

loss,

;

equal to the square of the current times the resistance of the

armature winding. For example, the resistance on a given armature is 72 = 0.1 ohm, and the total current supplied to the load and shunt-field winding is / = 150 amperes; then the volts drop in the armature is #d=/# = 150X0.1 = 15

and the watts

volts,

The watts

The

armature is W =EaI = 15 X 150 = 2,250 watts. TF=/ 2 fl = 150X150X0.1 =2,250.

loss in the

loss is also

losses in the

armature copjjer vary from about 4 per

cent of the capacity of the machine in 1-kw. units to 1.8 per cent for units of 1,000-kw. capacity. These losses vary as the

square of the current supplied by the armature and are practically zero at no load, being only those due to the shunt-

winding current, and at a maximum value at maximum The resistance of the armature circuit is usually considered as that of the armature windings, brushes, series-field windings if the machine is compound-wound, and the machine field

load.

leads

and terminals.

Eddy-Current Losses. In Chapter XI it was shown that when the armature core is revolved between the polepieces, it cuts the lines of force and therefore generates a voltage the same as the windings do. It was also shown that the current caused to circulate around in the core by this voltage, or eddy current as

called, created a pull that opposed the turning prime mover driving the generator, consequently

it is

effort of the

represented a direct loss of power. The eddy-current losses are usually combined with the hysteresis losses and are called the iron or core losses. Hysteresis Losses. The hysteresis losses are those which are due to the friction of the molecules, of the iron in the

ELECTRICAL MACHINERY

218

on each other as they align themselves with the the armature is revolved. This is illustrated in Figs. 235 to 237. In Chapter II it was explained armature

core,

lines of force

when

that a piece of iron acted as

if

each molecule was a magnet

3333 333333 333333 33333333

A 13333333333

FIG. 237

FIG. 236

FIG. 235

Shows how iron molecules in armature core are suparrange themselves when field poles are magnetized.

FIGS. 235 to 237.

posed to

having a north and a south pole, and that under normal conditions the molecules arrange themselves so that the N and S poles of one molecule were neutralized by the N and S poles of other molecules and thus form a neutral condition, as in Fig. 238. When the piece of iron is brought under the pole of a

3333333333333333 3333333333333333 333333333333333 FIG. 238

FIG. 238.

Supposed arrangement of the molecules when a piece of iron

FIG. 239.

Supposed arrangement of molecules when iron

is

not magnetized. is

magnetized.

magnet, this pole will attract the opposite pole of the molecules of the iron and cause them to be arranged in a systematic group, as in Fig. 239, thus producing an N pole at one end of

In the same way, when the machine are magnetized, they cause the molecules in the armature core to arrange themselves systematically as in Fig. 235. Now if the armature core is turned 90 deg. from the position in Fig. 235, as in Fig. 236, it the bar and an S pole at the other.

field poles of

an

electrical

LOSSES IN DIRECT-CURRENT MACHINERY will be seen that although the core as a

219

whole has revolved 90

deg. to the left as indicated by AB, the 'field magnets have held the molecules of the iron core in the same position in

For this to be possible the molecules have done what is equivalent to turning to the right 90 deg. If they had remained in a fixed position in the core, the condition that would exist is that in Fig. 237, from which it is seen that if each molecule turns 90 deg. to the right from the position in each case.

the figure, a condition exists in the core corresponding to that in Fig. 236. This is just what appears to be going on in the armature core all the time that it is revolving and the field poles are magnetized. As the armature revolves as a whole in one direction the molecules are revolving about their axis in

the opposite direction. The molecules revolve at the same rate as the armature core in a two-pole machine, or one revolution for one pair of poles. The latter statement conforms to the condition existing in all multipolar machines; that is, the molecules make a complete revolution about their axis for each

pair of poles in the machine. In a four-pole machine they would be revolving twice as fast as the armature, in a six-pole

machine three times as

fast, etc.

To cause the molecules to revolve about their axis requires a certain amount of power. The power that is expended in changing the position of the molecules is the hysteresis losses in the core. This, combined with the eddy-current losses, is called the core losses, and amounts to about 4 per cent in machines of 1-kw. capacity to about 1.2 per cent in 1,000-kw. machines. electrical

If an attempt is made to turn the armature of an machine by hand, with the field poles dead, it should

turn very easily, but when the field poles are magnetized, it will be found that a greater effort must be developed to turn the armature. This increased effort required under the latter condition is due almost entirely to hysteresis, or in other words, to rotating the molecules of the core, and when the machine is in operation represents a loss of power.

Total Losses in an Electric Machine. full load in a 1-kw.

machine amount

The

total losses at

to about 20

per cent of

ELECTRICAL MACHINERY

220

the output, while in the 1,000-kw. machine they are about 4.5 In other words, when a 1-kw. machine is to 5 per cent. delivering

it,

it will require about 1.2 a 1,000-kw. machine is delivering its

load (1-kw.)

its full-rated

kw. to drive

and when

full-rated load (1,000 kw.)

drive

The

ratio of the output of a given

and

called the efficiency .

thus: Per cent efficiency

=

will require about 1,050 kw. to

it

it.

input ~

X

is

machine

to the input is

usually expressed in a percentage,

= output X 100

.

-.

per cent efficiency " .,,-, and input 100

,

from which, output output

X

100 .

,

.

per cent efficiency

For example, a given generator requires 50 hp. to drive it when supplying 32 kw. to a lighting system. Find the percentage of efficiency that the machine is operating at.

The input

in this case

is

Kilowatts

horsepower,

= hp.X74Q

50X746 37.3;

_

1,000

1,000

then

Per cent

efficiency

= outputX

100

32X 100 =

input

That

is,

37.3

86 per cent approximately.

only 86 per cent of the power supplied to the machine work in the lighting system.

is

available

for doing useful

FIG. 240.

Diagram of shunt-generator connected

to load.

A

given shunt generator, Fig. 240, when supplying a constant load, The voltage at the armature terminals requires 175 hp. to drive it. under this condition is 115. The core losses of this machine amount to

The field-circuit resistance 4.5 hp. losses amount to 3,500 watts, and to

is 11.5 ohms, the armature-copper overcome the friction of the machine

LOSSES IN DIRECT-CURRENT MACHINERY Find the percentage of

requires 1.5 hp. is

221

which the machine

efficiency at

operating.

The input in this case is 175 hp., and the output will be equal to the input minus the losses in the machine. The field current ii is equal to 1

77T

the volts divided by the field-circuit resistance

r\,

or ii=

-I

F

= 10

= n

11.5

The field watts W\ =Eii = 115X 10 = 1,150. The core losses are 4.5 hp., or 4.5X746 = 3,357 watts; 1.5 hp., or 1.5X746 = 1.119 watts is expended in overcoming the friction of the machine, and the armature amperes.

losses are 3,500 watts.

Then the

total watts loss

+3,500 = 9,126 watts. The horsepower input

is

is

1,150+3,357+1,119

175, or

175X 746 = 130,550

watts.

Output equals the difference between input and the. losses in the machine, or in this case 130,550-9,126 = 121,424 watts. Per cent

efficiency

= vutputXlQQ = 121,424 = 93 :

input

per cent.

130,550

Find at what value a circuit-breaker should be set, or what size fuses should be used, to protect a 75-kw. 120-volt generator if the load is to be limited to 120 per cent of normal.

The normal load in watts would be the normal load in kilowatts multiplied by 1,000 (since there are 1,000 watts in a kilowatt) or 75 1,000 = 75,000 watts.

X

The normal current would be the normal load voltage of the generator, or

75,000

= 625

in

amperes.

watts divided by the If

the load

is

to be

limited to 120 per cent of the normal value, the current would be 1.2 times the normal current, or 1.2X625 = 750 amperes. The circuit-breaker

would therefore have to be

set for

750 amperes, or fuses of that capacity

be used.

A generator supplies 750 amperes at 130 volts while operating at 89 per cent efficiency. How many horsepower would be required to drive it?

The load supplied by the generator, expressed in watts, would be the current which it delivers multiplied by the voltage, or 750X130 =97,500 watts. If it operates at 89 per cent efficiency, 97,500 watts must be 89 per cent of the power required to drive it. the power required to drive the generator, we have 0.89 from which

= 109,550 W = 97 500 89 '-

watts.

If

W

represents

W = 97,500 watts,

Since there are 746 watts to

.

one horsepower, the power required to drive the generator in watts divided

by

746, or

109,550 -=

746

147 hp. (nearly).

is

the power

CHAPTER XVI DIRECT-CURRENT MOTORS Fundamental Principle of Electric Motor. Figs. 139 to Chapter X, explains the fundamental principles of the

145,

electric motor, therefore

taking

up

this chapter.

should be carefully studied before has been shown that when a cur-

It

rent flows through a conductor in a magnetic field it will tend to cause the conductor to move at right angles to the lines of force.

The

depend upon the direction of the

will

move in force and the

direction that the conductor will tend to lines of

direction of the current in the conductor.

This relation

is

explained by the rule Figs. 143 and 144, Chapter X. Consider an armature such as is used in a shunt generator, Fig. 241, to have current passed through it after the field coils have been connected to the mains; then there will be a

number field, all

of current-carrying conductors situated in a magnetic of which will tend to be moved across the path of the

If the forces impelling them are all in the same directhey will cause the armature core on which they are mounted to revolve. The direction of the force on any con-

field.

tion,

ductor can, as previously described, be determined by the The relative finger rule, as illustrated in Figs. 143 and 144. direction of the magnetic field and current will depend on the position of the brushes on the commutator. It will be found that, theoretically, the brushes should be

on the neutral axis

for proper operation, exactly as was found in the case of the

generator.

In Fig. 241 the direction of the current in each conductor above the neutral axis of those below

is

away from

the observer, and in each toward the observer.

this axis the current is

222

DIRECT-CURRENT MOTORS The

223

direction of the field in each case

is upward, consequently will tend to move to axis neutral the above conductor every the right and every one below this axis will tend to move to the left, the result being that the armature will be made to

revolve in a clockwise direction.

FIG.

Effects of

be

moved

241.

Diagram of

direct-current motor.

Changing Brush Position. Should the brushes shown in Fig. 2'42, the armature will

to the position

refuse to rotate, for then the direction of the current in the armature conductors between a and 6 and those between a and d will be toward the "observer. One set of these conductors will therefore tend to revolve the armature in a counterclockwise direction

and the other

in a clockwise direction.

current in conductors between

c

and

&

and

Likewise the

in those between

and d is down through the plane of the paper. Hence we have conductors between a and & and between c and d producing

c

counterclockwise rotation and those between &

and

c

and

between d and a opposing them with an equal force in the opposite direction; consequently there can be no motion.

ELECTRICAL MACHINERY

224

For intermediate positions of the brushes, the total number of conductors tending to move in one direction will be greater than the total number tending to move in the other direction under the same

pole,

hence there will be rotation in the direc-

tion of the greater force. The nearer the brushes are to the neutral axis the greater the force becomes until exactly at the

neutral axis the force

FIG.

242.

Same

is

at a

as Fig. 241;

maximum.

except brushes are shifted 90 degrees.

From

the foregoing it is evident that a shunt generator can be used as a motor. The same holds true for all the other

types of generators, and as a matter of fact, whether a dynamoelectric machine is called a generator or a motor depends largely on the name-plate.

name-plate and replace

We might take a generator,

remove

with one specifying a slightly different rating, and would have a motor of the same type. If the generator were a compound one, the motor would be a

its

it

compound motor, etc. Another Theory of Motor Operation. ther

it

Before going fur-

will be of interest to discuss a theory of

motor operation

DIRECT-CURRENT MOTORS which

is

225

from the one already given, but which exwell as the foregoing how the motor operates.

different

plains quite as In Fig. 243, the current flowing through the

upper half of the

N

in a direction to produce an pole on the lefthand side, and an S pole on the right-hand side. The same Therefore we have an is true of the current in the lower half.

armature

N

is

N

pole in the lower half, upper half opposed by an resultant and S poles for S the consequently likewise, poles, are produced on the armature as shown in Fig. 243. With in the

N

FIG. 243.

Same

as 241, except armature poles are indicated.

the field poles magnetized, the N pole of the armature will be attracted to the S pole of the field frame and the S pole of the

armature to the N main pole, therefore it will begin to turn in a clockwise direction. So long as the brushes rest on commutator bars a and c, the direction of the current through the windings remains unchanged, hence the N and S poles remain fixed in the ring and turn with it. Fig. 244 shows their position when the commutator bars a

and

c

have

left the

brushes and bars

~b

and d have made contact

ELECTRICAL MACHINERY

226 with them.

At

the instant

rent in coils 1 and 3

is

when

this

change occurs, the cur-

reversed and the

N

and S

poles of the

armature are instantaneously transferred backward through a space equal to that occupied by one armature coil so that they are again opposite the brushes as in Fig. 243. In this we have the main poles tending to pull opposite poles of the armature up to them, but as fast as the armature

moves up one

Same

FIG. 244.

the current

is

reversed in the coil by the

as 243, with armature

moved one commutator segment.

coil

commutator action, and as a result armature poles are maintained in a position at the brushes, as the armature rotates. Although ring armatures have been used in the foregoing description for the sake of clearness, the theories advanced and S poles are apply equally to drum-type armatures, since

N

The first formed precisely as in ring armatures. in a a conductor motion of on the theory presented, that based in

them

due to the current being passed through it, is the more fundamental of the two, but the second one, that based on the attraction of magnetic poles, is extremely conmagnetic

field

DIRECT-CURRENT MOTORS venient and

is

227

a considerable aid in coming to a thorough

understanding of the principles involved.

Both Motor and Generator Action Present in all DynamoHaving given an armature situated in a in a field as dynamo-electric machine, the machine magnetic becomes a generator when the armature is caused to revolve by

Electric Machines.

applying mechanical power to it, whereas when electrical power is supplied to it, the machine becomes a motor and

FIG. 245.

Indicates both motor and generator action in same directcurrent machine.

delivers mechanical power.

In both the cases a number of con-

ductors are carrying currents and they are moving across a magnetic field. The question then arises, what is the difference

which causes the machine to act as a generator in one case and as a motor in the other if the conditions in both cases are the same ? Briefly stated, it is that the two actions are opposite to each other and that in one case the generator action is greater than the motor action, and in the other the motor action exceeds the generator action. That is, both actions are present

ELECTRICAL MACHINERY

228

whether the machine be operating as a generator or as a motor, but the one is greater than the other, depending on whether it is being used to convert mechanical into electrical energy or vice versa.

Consider how these actions combine to give the results obtained. Suppose we have an armature situated in a magnetic field, as in Fig. 245, and that the direction of the current in the armature winding

and the direction of the

246.

is

as

FIG. 247

FIG. 246

FIG.

flux

rule

for

generator. Right-hand Left-hand rule for motor.

FIG. 247.

Unless the direction in which the armature was

indicated.

known, it would be impossible to say whether it was being driven as a generator or was running as a motor. It might be operating as either. As a generator it would have to be driven in the direction of the arrow A if the current is to be in the direction shown, as may be determined from the righthand rule illustrated in Fig. 246. As a motor it would revolve in the opposite direction if the direction of the current was made the same, as shown in the left-hand rule illustrated in revolving

is

DIRECT-CURRENT MOTORS

229

Fig. 247. It is therefore apparent that, as previously stated, the motor and generator actions are always opposed.

Counter-Electromotive Force.

Suppose the machine

to be

operating as a generator. When it has a load connected to it, the current passing through the armature tends to make all the conductors of the winding move across the field in the direction B, and in order that this force shall not reduce the

speed at which the engine or other prime mover is shoving them in the direction A, the prime mover must exert a pull equal to the drag of the conductors. Consider, now, that the machine is operating as a motor. It When the will, of course, be revolving in the direction B.

armature conductors move in this direction through the field an e.m.f. would be generated in opposition to the applied voltage, illustrating in another way how the motor and generator actions oppose each other. The e.m.f. which is generated

by the conductors of the motor in opposition

to the applied

called the counter-electromotive force of the motor voltage it is abbreviated counter-e.m.f., or c.-e.m.f. and is also often is

;

referred to as back-e.m.f., the words counter cating that it is in opposition.

and back

indi-

Since the armature of the motor generates an em.f. in opposition to the voltage applied to it, the current which can flow through it is equal to the difference between the applied voltage and the counter-e.m.f. divided by the resistance of the

That

armature.

is,

Applied voltage counter-e.m.f. r-r armature resistance I

=

= armature

,

current, that

Tft

=

,

JK

volts, e the

where 7 equals the armature current, counter-electromotive force, and

ance in ohms.

The

R

.

is,

E the applied

armature

resist-

counter-e.m.f. of a motor will always be less

than the applied voltage, for if the two were to become equal to each other no current could flow, and the armature would The moment the armature began to slow down, a very stop. interesting action

would take

place, as

shown

How Motor Operates when Loaded.

in the following

:

In the study of gen-

ELECTRICAL MACHINERY

230 erators

it

was found that the

em.f. generated

by conductors

moving across a magnetic field is proportional to the rate at which the lines of force are cut by the conductors. If the speed

is

decreased while the field strength

is

e.m.f. will be decreased because the total

kept constant, the

number

of lines of

force passed in one second will be less than before. It follows that the instant the armature of the motor slows down, the em.f. which its conductors generate will be decreased. That is,

its

counter-e.m.f would become less than the applied voltage, which a current would flow, and as soon as .

in consequence of

the current became great enough to keep the armature revolving, there would be no further reduction in speed the arma;

ture would continue to revolve at whatever speed it had dropped to. If the motor is running idle, only a small cur-

rent will be required to keep it running, merely enough to overcome friction and other internal losses. The applied volttage multiplied by the current flowing at no load will give the watts used up in the motor to overcome its losses. Were a load now to be placed on the motor it would of course tend to retard The speed would again begin to drop and would its motion. continue to do so until the counter-e.m.f. had been reduced to such a value as would allow sufficient current to flow to over.corne the load imposed on the motor. The product of the volt-

age and current would now be the watts required to drive the ioad plus the internal losses of the motor. Provided the supply voltage remained constant we would find a certain definite speed for each load, and the greater the load the lower the speed. The amount of change in speed for change in load differs greatly in the different types of motors. suffers the least change, the series motor the the compound motor being in between the two. It is well most, to note here that the difference between the applied voltage

The shunt motor

and the counter-e.m.f.

is

never very great.

An

example will

illustrate the reason for this. If

a 5-hp. 110-volt motor whose armature resistance is 0.1 full load, it would require about 5X880 = 4,400 watts.

carrying size

motors require about 880 watts per horsepower output.

ohm

is

Small-

The current

DIRECT-CURRENT MOTORS

231

required equals watts divided by voltage, or 4,400-r- 110=44 amperes. The voltage drop in the armature would be 44 amperes multiplied by

ohm, or 4.4 volts, which would be the difference between the applied voltage and the counter-e.m.f., and the value of the latter would therefore be 110 volts minus 4.4 volts, or 105.6 volts. In this case, the drop 0.1

Any material decrease only about 4 per cent for full load. would allow an enormous current to flow and would

in voltage is

in the counter-e.m.f.

result in the operation of the safety devices protecting the motor and the circuit to which it is connected. For instance, if the counter-e.m.f.

should drop to 80 volts, the difference between it and the applied voltage would be 110 volts minus 80 volts, or 30 volts, and the current taken by the armature would be this difference, divided by the armature resistance, or 30 volts divided

by

0.1

ohm, equals 300 amperes, which

is

seven times

the full-load current.

How

Speed of Motor

May Be

Adjusted.

It

has been said

that the speed of a motor adjusts itself to the value at which the correct counter-e.m.f. for the particular load on it will be

generated; that is, the counter-e.m.f. is the controlling factor in the automatic regulation of the motor. The counter-e.m.f.,

however, is affected by the strength of the magnetic field, which leads to another feature in the operation of the machine. Suppose a shunt motor, for example, to be revolving at a certain speed while driving a load. If the field strength were to be it would be if we were to decrease the field current any reason, then the counter-e.m.f. generated would drop in proportion. A drop in the counter-e.m.f., however, would cause the motor to take a greater current than required by the load, consequently the motor would speed up. It should be remem-

reduced, as

for

'

bered, however, that the field flux is less than it was, but the percentage increase in armature current far exceeds the

percentage decrease in flux, and the armature does speed up. up, the counter-e.m.f. again increases, thereby decreasing the armature current until the point is reached at which the current flowing is again just sufficient to take care

Upon speeding

of the load.

Since the load

more power

is

being driven at a greater speed,

it, and consequently, the value of armature current will be somewhat higher than that previously observed.

it

will require

to drive

new

When

the field current

is

increased, the consequences are

ELECTRICAL MACHINERY

232

the converse of those described and the speed of the motor is As before, assume that while a shunt motor is run-

reduced.

ning under a given

set

of conditions, the field current is is thereby increased and the

increased, the counter-e.m.f. armature current decreased.

The motor then takes an amount

of current insufficient to drive the load at the original speed

and

is

e.m.f.

compelled to slow down. Upon doing so, the counterdecreases, allowing the armature current to increase.

The speed

will decrease until the

armature current has

tained a value sufficiently great to take care of the load.

new current

will

be somewhat

less

than that previously

observed since the speed has been somewhat reduced. The fact that the speed of a motor increases when the current decreases and decreases

field

the field current in-

an extremely important one, since on it depends the method of control of adjustable-speed motors.

creases, is

chief

when

at-

The

CHAPTER XVII TYPES OF DIRECT-CURRENT MOTORS Five Types of Motors.

Direct-Current motors

may

be

divided into five general classes according to their electrical characteristics

;

namely,

and compensating

series,

types.

In

shunt and compound, interpole all these types the armatures

be the same, the essential difference being in the fieldpole windings and the way the latter are connected to the armature. The simplest type is the plain series motor, which derives its name from the fact that the armature and field

may

windings are connected in

FIG. 248.

series.

Since the armature and

field

Connection diagram for series motor and starting box.

windings are connected in series, the latter passes all the armature current. Consequently, the wire in the field coils must be large enough to carry the total load current without excessive heating or voltage drop.

For

this reason the field

a series motor are usually wound with a comparatively The small number of turns of relatively large-sized wire. coils of

connections of a series motor to an ordinary type of manual It will be noticed that starting box are shown in Fig. 248. 233

ELECTRICAL MACHINERY

234

but one circuit through the starting box, armature and field windings, the armature and field windings being connected in series. Where motors of this type are of 3.5 hp. in size and above, they can generally be determined by inspection, there

is

by noting if the machine has only main poles, as in the figure, and if the field and armature leads are the same size. A test can be made with a lamp and circuit having machine voltage. If the motor is of the series. type, the lamp should light up to full brilliancy when connected across either the field or armature terminals excepting in the case of very small machines. Torque of Series Motor Varies as the Square of the Current, is

that

One it

of the serious disadvantages of a series motor

will race if its load

is

removed and

unless disconnected from the line.

the motor, the speed will

vary inversely

increased the speed will decrease,

will destroy itself

If the load

and

if

that

is

changed on

is, if

the load

is

the load

is

decreased

consequently, the motor is not adapted For this reason the motor is used to constant-speed service. to a very limited extent excepting in cases where it is con-

the speed will increase

tinuously under the control of an operator; for example, on The motor has a characteristic cranes and railway service. that makes it well adapted to these two -classes of service;

namely, within certain limits its torque (turning effort) varies as the square of the current that is, if the current taken from the line is increased by 2, the torque will be increased four ;

Why

this is so may be readily understood by a conwhat happens in the motor when the value of the current is doubled in the motor's windings. For example, if the motor takes 10 amperes from the line, this current passes through both the armature and field windings and produces a

times.

sideration of

number of lines of force per square inch in the field which is reacted upon by the current passing through poles, the armature conductors and causes the armature to develop a given turning effort. If the current was doubled in the armacertain

remaining constant, the turning effort would be doubled. On the other hand, if the value of the current in the armature was maintained constant and the lines of

ture, the lines of force

TYPES OF DIRECT-CURRENT MOTORS force

doubled in the Since the

doubled.

235

pole, the torque would also be and armature windings are connected

field

field

is increased to 20 amperes, then the current will be doubled not only in the armature 's conductors, but also through the field coil. The latter will increase the

in series, if the current

lines of force

field pole by 2. Consequently, when the current in the armature conductors

from the

we have two times

acting upon two times the flux from the polepieces, this will give four times the turning effort developed by 10 amperes.

Of course

this holds true only

up

to the point

where the

approach saturation. Then the increase in current is not so marked.

magnetic circuit begins to

increase in torque for an This characteristic of the torque increasing as the square of the current, makes the motor well adapted to work that re-

quires frequent starting under heavy load.:, as in railway and crane work, and it is in these two applications that the series

motor finds

its

greatest application.

Shunt-type Motors. The plain shunt type of motor is in general appearance very much like the series machine, and in the smaller sizes it would be somewhat difficult to distinguish one from the other by a casual inspection. The shunt machine gets its name from having its field coils connected in parallel,

or in shunt as

it is

sometimes called, with the armature.

The

connections for a simple shunt-type motor and manual-type starting box are shown in Fig. 249. Instead of one circuit, as in the series motor, there are two in 'the shunt machine;

namely, one through the armature and another through the field winding. The field coils, Fig. 249, are connected directly across the line,

therefore must be

wound with a wire

of comparatively

small cross-section, so that the exciting current will be maintained at a value consistent with high efficiency, and also to

prevent the

marked

coils

from overheating.

In this we find the most

distinction between the series-

machines.

In the series machine the

number

and the shunt-type

field coils

are

wound with

of turns of large-sized wire, low have a while in the shunt machine resistance, consequently

a comparatively small

ELECTRICAL MACHINERY

236

wound with a large number of turns of comparatively small wire and have a high resistance. In largesized machines, 3.5 hp. and up, the motors can be distinguished

the field coils are

size of the field leads, they being smaller than the armature leads on the shunt machines and the same size on the

by the

In the small machines, where the field and armature leads are the same size on both types, one type may be distinguished from the other by testing through the field coil with a lamp. When tested on the circuit of the machine's rated voltage a lamp connected in series with the field coil of

series machines.

FIG.

249.

Connection diagram of shunt motor and starting box.

the shunt machine will glow dimly, but on a series machine the

lamp will light brightly. Torque of Shunt Motor Varies as the Armature Current. Since the field winding of a shunt motor is connected across the line, the magnetic density in the field poles will be pracTherefore the torque tically constant irrespective of the load. of a shunt motor varies directly as the current in the armature, and not as the square of the current, as in the series machine.

In other words, if the current through the armature of a shunt motor is increased by two, the torque is also increased two Also, since the strength of the field pole is practically constant, the speed of the motor will be practically constant from no load to full load. In practice the variation is only

times.

about 5 to 10 per cent from no load to full load, the speed being higher at no load than at full load. Therefore, in a

TYPES OF DIRECT-CURRENT MOTORS

237

shunt motor we have a machine the torque of which varies directly as the current in the armature and whose speed is These two features make this type of practically constant. to a motor applicable large number of drives where the startnot severe and where a constant speed is ing conditions are required over a wide range of load. Compound-type Motors. There

are

many

drives

that

require practically a constant speed after the motor has been accelerated and where strong starting torque is required. In other words, a motor is required that has speed characteristics

and the torque characteristics of a series To meet such requirements, what is known as a

of a shunt machine

machine.

is used, which is nothing more nor less than a shunt machine with the addition of a low-resistance winding

compound motor on the

field pole,

connected in series with the armature.

The

connections for a simple type of compound motor to be started by a manual-type starting box are shown in Fig. 250. It will

FIG.

250.

Connection diagram of compound motor and starting box.

be seen that the shunt winding, Sh is connected across the line, the same as in Fig. 249, and the series winding 8 is connected in series with the armature, as in Fig. 248. The same starting ,

box

is

used on both the shunt- and compound-wound motors. motors are used on elevator service, the

When compound controller

is

usually arranged so that the series winding

circuit only while the

motor

is

coming up

to speed.

is

in

After the

ELECTRICAL MACHINERY

238

load has been accelerated, the series winding is cut out of circuit automatically, thus converting the motor into a shunt

machine during the running period.

In

this

made to act as a generator when the car down motion under heavy load to prevent

is

way

the motor

traveling in the the machine from

is

Cutting out the series winding during the running period also tends to make the motor run at a more constant racing.

speed under variable loads. Although a compound motor will not race as will a series machine under light loads, nevertheless its speed machine.

tions

is

not as constant under variable loads as a shunt

With

the average compound motor the speed variato full load is in the neighborhood of 15 to

from no load

20 per cent. This, however, will depend upon the series The turns (turns X amperes) on the polepieces. ampere in of the number turns the series greater ampere winding the greater will be the starting torque, as also the variation of the

motor speed. There are two ways of connecting compound motors so that the series and shunt-field windings have the same polarity, and also so that the two windings will have opposite polarities. Where the two sets of field coils are connected so as to have the same polarity, the machine is called a cumulative compound motor; that is, the resultant field strength is the cumulative effect produced by the two windings. When the two windings are so connected that they oppose, the machine is called a differential compound motor, since the resultant field strength is the difference between that produced by the shunt- and the

series-neld windings.

The simple series, shunt and compound types, of directcurrent motors cover in a general way motors for use on directcurrent circuits. There are a number of other types, but they are only modifications of the three types discussed in the foregoing. Effects of

In all direct-current maan electromagnet with its north and south poles just as the field poles are, as was shown in Chapter XVI. Theoretically, the armature poles should be located mid-

Armature Reaction.

chines the armature

is

TYPES OF DIRECT-CURRENT MOTORS

239

way between

the polepieces, but owing to the reaction between the two sets of poles the armature's neutral is caused to shift against the direction of rotation in a motor and with the direc-

tion of rotation in a generator. This shifting of the armature 's neutral is one of the causes of sparking at the brushes in the

simple types of direct-current machines and will be made clear in the following discussion In Fig. 251 is shown a schematic diagram of a direct:

current machine having a ring-wound armature. If current is sent through the field coils, a magnetic field will be set up

which will be uniformly distributed as in the

figure.

Leaving

the field poles dead and passing current through the armature only with the brushes set midway between the polepieces, will

cause magnetic poles to be produced in the armature midway between the main poles, as indicated in Fig. 252. The lines

shown in the The armature poles are really the resultant of two north and the two south poles. In following the direction of force will be distributed similarly to that figure.

two halves of the armature winding and an S pole are produced in the upper half of the armature, likewise in the lower half. The N poles are produced on the right-hand side and the S poles on the left. This brings of the current through the

an

N

two

like poles together at each brush; these poles repel each other, with the result that the lines of force tend to distribute

as in the figure.

Comparing Fig. 251 with Fig. 252, it will be seen that the field set up by the armature is at right angles to that

magnetic

of the field poles. When current flows through the armature and field coils at the same time, as in Fig. 253, the pole of

N

the armature will repel that of the field poles and attract the S pole. Likewise the S pole of the armature will repel the

main S pole and

attract the

N

pole, with the result that, in-

stead of the lines of force being uniformly distributed as in Fig. 251, the field will be distorted as in Fig. 253 that is, the ;

pushed away from one pole tip over onto the other. This action will vary with the load on the machine. If the load is small, the current in the armature is at a low value,

lines of force are

ELECTRICAL MACHINERY

240

the field poles of the armature will be weak and have very little effect upon the main poles, consequently the line of force from the latter will be nearly uniformly distributed, as in Fig.

However, as the load comes on, the magnetic field of the armature will increase and its effect to distort the lines of force from the main poles will be increased. Therefore, with a vary251.

FlG. 251

FIG. 252

FIG. 254

FIG. 253

FIGS. 251 to 254.

Shows magnetic

flux

distribution in

direct-current

machine.

ing load on the motor the distribution of the field from the main poles will be changing. With the flux uniformly distributed, as in Fig. 251, the correct position of the brushes

would be

as indicated in the figure, or

what

is

known

as the

neutral point. When the motor is loaded as in Fig. 253, the neutral point has shifted to some position such as indicated by

TYPES OF DIRECT-CURRENT MOTORS

241

ab due to the distorting effect of the armature poles. Consequently, the correct position of the brushes is not as shown in the figure, but on the line ab. From this it is seen that on the

simple type of direct-current motor the neutral point shifts with the load, and this is one of the reasons for sparking at the brushes with changing loads on the machine.

To prevent

Commiitating-pole Machines.

this shifting of

the neutral point or to at least hold a magnetic field at this position where the brushes are located, to assist in commutat-

ing an additional set of poles is frequently used in modern direct-current machines. These consist of a small pole called an interpole or commutating pole, located in between the

main poles with a low-resistance winding connected in series with the armature as in Fig. 254. These poles are magnetized the same polarity as the armature poles, as indicated, and if of the same strength they will neutralize the effect of the armature poles and the neutral will remain in a fixed position. Since they are connected in series with the armature, the strength of their magnetic fields will vary as those of the armature.

At

light loads the current in the

armature and interpole

windings will be small, consequently the field poles of each will be weak. As the load increases, the current through the

and interpole windings increases, likewise the strength of the field poles of both consequently, if the interpoles are properly designed, they will neutralize the effect of armature

;

the armature pole at all loads

from no

load* to full load,

making

possible to obtain sparkless commutation throughout the whole load range of the motor with the brushes in a fixed posiit

tion.

Number of Interpoles Used. An interpole for each main pole in the motor may be used, as in Fig. 255, B being the main poles and the interpoles. However, to reduce the cost

A

of construction, sometimes only one interpole pair of main poles, Fig. 256. In this case it

is

used for each

is

essential that

the interpoles be so located that they are all the same polarity that is, in Fig. 256 they would either be two north poles or two ;

south poles.

But the

best practice

is

to use

an interpole for

ELECTRICAL MACHINERY

242

each main pole, and this is adhered to in a large percentage of cases, especially on medium-sized and large motors. Interpoles are used on all three, types of direct-current although as far as the series, shunt and compound

machines

external connections are concerned, the addition of the interpoles to the machine does not in any way complicate the connections over those for the simple types without interpoles.

FIG. 255.

In

main pole. Four-pole interpole motor; one interpole for each

Fig. 257 are shown the connections for a four-pole series motor with interpoles, to be started from a manual-type starting box. These are the same as for the simple type of machine. On the

the shunt-interpole type only four leads are brought outside made are starter motor and the connections to a manual-type If it was desired to reverse the motor, the as in Fig. 258.

connection at the armature terminals would be interchanged,

TYPES OF DIRECT-CURRENT MOTORS

243

which would reverse the direction of the current through the armature and the interpole windings. This is exactly as it the direction of the current through the reversed, the polarity is reversed, and if the inter-

should be, since

armature

is

if

same polarity as the armature, the current through this winding must also be reversed. poles are to remain the

FIG.

256.

Four-pole interpole motor; one interpole for each pair of

main

poles.

In the compound-wound interpole machine the interpole is connected in the circuit between the armature and

winding

series winding.

The connections

to a

manual-type starter are

given in Fig. 259. It will be seen that, externally, these are the same as for a compound machine without interpoles. To reverse the direction of rotation of the motor, all that is necessary is to interchange the connections on the interpole and

armature terminals outside the machine.

This will reverse the

ELECTRICAL MACHINERY

244

direction of the current through the armature and interpole windings as in the shunt machine. From the foregoing it is

FIG. 257.

Connection diagram for series-interpole motor.

FIG. 258.

Connection diagram for shunt-interpole motor.

FIG. 259.

Connection diagram for compound-interpole motor.

evident that the connections of interpole machines to their starting device are the same as for those without interpoles.

TYPES OF DIRECT-CURRENT MOTORS

FlG. 260.

FIG. 261.

245

Field frame for compensated type motor.

Field frame and windings for compensated-type motor.

ELECTRICAL MACHINERY

246

Motors with Compensating Windings. In some applicawhere the service conditions are very severe, interpole

tions

windings are not sufficient to give sparkless commutation. To meet these conditions the pole faces are slotted as in Fig. 260

and windings, known

as compensating windings, placed in

these slots, as in Fig. 261. is

the interpole winding.

This winding is of low resistance, as The current in the compensating

windings flows in an opposite direction to that in the adjacent armature conductors, consequently if the compensating winding has the same number of conductors as the armature then the former will completely neutralize the effects of the latter.

The armature may be looked at as an electromagnet with its and S poles between the main poles, and the compensating

N

coils a second winding on this electromagnet wound in opposition to the armature winding. Since the windings are in if of are opposition, they equal number of turns, one will

neutralize the effect of the other

through them, and the pole faces does.

this is

when current

is

passing

what the compensating windings in

CHAPTER XVIII STARTING RHEOSTATS AND CONNECTIONS TO DIRECT-CURRENT MOTORS Starting Resistance. Unlike a coil of wire in which the current is limited by the resistance of the circuit, the current

armature of a direct-current motor

in the

is

limited almost

entirely by the counter-electromotive force generated in the conductors as explained in Chapter XVI. When the motor is

running, the counter-e.m.f generated in the armature is almost equal to the applied voltage at standstill it is zero. The resist.

;

low, therefore a resistance must be connected in series with it at starting to limit the current to a

ance of the armature

is

safe value.

The

resistance connected in series with the armature

is

usually of a value that will limit the starting current to about 125 per cent of that at full load. That is, if the full-load

current of a motor is 20 amperes, the starting resistance will be of such a value as to limit the armature current to 20

X

=25

When

the armature starts to revolve, a counter-e.m.f. is generated and the current is reduced until the machine reaches a certain speed, which will be fixed by the 1.25

load.

At

amperes.

this point, if a section of the resistance is cut out, the

current will be again increased, and this will raise the speed of the armature, causing the counter-e.m.f. to increase, and de-

by the load on the motor. Action Taking Place when Motor Is Started. Just what takes place when the motor is starting will probably be better understood by an example crease the current to a value fixed

:

Assume a condition as shown in Fig. 262; the armature resistance ohm, and the starting resistance ^2=8.5 ohms, a total resistance of 0.5 +8.5 = 9 ohms. With 225 volts impressed on the circuit at the instant Ri =0.5

247

ELECTRICAL MACHINERY

248

of closing the switch the current / that will flow will equal the effective That is. e divided by the ohmic resistance R. voltage

E

/

=

E = 225 = 25 e

amperes.

Note that the effective voltage is the difference between the line volts and the counter-e.m.f. When the motor is standing still, the countere.m.f.

is

zero,

consequently the total line e.m.f.

is effective.

The

pressure E\ across the motor terminals will equal the armature Hence Ei=Ri I =0.5X25 = 12.5 resistance Ri times the total current 7. as shown.

volts,

The

difference

between the

line pressure

=EEi = 225

used up across the resistance, or EI E-2 is also equal to the starting resistance is

# = #27= 8.5X25 = 212.5 2

R

12.5

E

and EI

= 212.5

volts.

times the current 7; that

is,

volts in either case.

//

*z-W\

i

t*K5 F.G. 262

FIGS. 262 and 263.

FIG. 263

Diagram of armature and

starting resistance.

As the motor speeds

up, a back pressure (counter-e.m.f.) is generated armature, which decreases the current. Assume that the motor attains a speed that will cause a counter-e.m.f. e of 90 volts to be generated in the armature. The effective pressure E e will equal the line in the

minus the back pressure e, or 22590 = 135 volts. The current flowing through the armature under this condition equals the line voltage E minus the counter-e.m.f. e divided by the total resistance R, or volts

E

7i,

E-e 225-90 = 15 R

9

amperes,

as indicated in Fig. 263. The volts drop E* across the starting resistance is product of the starting resistance 2 and the current 7i

R

= 8.5X15 = 127.5

volts,

equal to the that is, E 2 = R*I\

#!=# -#2 = 225 -127.5 = 97.5

volts

volt drop across the armature due to resistance equals 0.5X15 equal to the armature resistance Ri times the current

across the armature. is

which leaves

now ;

The

h

STARTING RHEOSTATS AND CONNECTIONS = 7.5

volts.

It will

249

now be

seen that this value (7.5 volts) plus the volts) equals 97.5 volts, the new value calacross the armature, as shown in the figure.

assumed counter-e.m.f. (90 culated for the drop From the foregoing

it will be seen that as the motor speeds up the current decreases and the pressure increases across the armature terThis continues to a point where the machine is just taking minals. current enough from the line to carry the load, or, as assumed in this

case, 15

Any

amperes.

further increase in speed would be followed by a and a decrease in the current. If

further increase in the counter-e.m.f.

the motor was starting under no load, it would run up to about full speed and the volts drop across the armature would be almost equal to the On moving the starting-box arm upon the second point, line pressure. the resistance will be reduced, say to 6 ohms, as shown in Fig. 264; now the ohmic resistance of the circuit equals 6+0.5 = 6.5 ohms. When the arm was moved upon the second point of the starting resistance, the effective pressure equaled 135 volts;

therefore, at this instant the current

will increase to

/2

=

E = 135 = 20.77 R 6.5 e

amperes approximately.

Under the new condition the voltage drop across the starting resistance will be #2 =#2/2 = 6X20.77 = 124.6 volts, and the drop across the armature due to the resistance equals #1/2=0.5X20.77 = 10.4 volts. The total voltage impressed on the armature equals that due to resistance

The values plus the counter-e.m.f. or in this case 10.4+90 = 100.4 volts. are shown on the figure and total up to the line volts, which is as it should be.

This increase of potential and current to the armature will cause an speed, which in turn will raise the counter-e.m.f. and

increase in the

again decrease the current to a value where the torque will be just sufficient to carry the load. Assume that the current decreases to /3

= 16 amperes. Then, the = 6X16 = 96 volts, and

#2/2

difference

shown

between

this

volts drop across the resistance will equal the drop across the armature will be the

and the

line voltage or

22596 = 129

volts, as

The

resistance of the armature will cause a drop of #1/3 = 0.5X16 = 8 volts, and the difference between this and the total = 121 volts. drop equals the counter-e.m.f. or in Fig. 265.

1298

The foregoing shows how the starting resistance limits the current to the armature and how the e.m.f. gradually increases at the

motor terminal as the speed

increases.

When

the start-

cut out, full line pressure will be applied to the armature terminal, the motor will come up to full speed, and the counter-e.m.f. will be almost equal to the line volts.

ing resistance

is all

ELECTRICAL MACHINERY

250

The

difference will equal the

armature resistance times the

current.

Starting-Current Curve. If the points are taken on a straight line, as in Fig. 260 to represent the points on the starting box, and at these points vertical lines are drawn to represent the amperes to scale, a curve may be obtained that For example when the will represent the starting current.

starting-box arm was placed on the first point, the current rose to 25 amepers, as shown by the vertical line drawn from point

As the motor increased in speed, the current decreased to 15 amperes before the starting-rheostat arm was placed on

1.

FIG. 264

FIGS.

FIG. 266.

point

2.

FIG. 266

FIG. 265

264 and 265.

Same

as

262, but with cut out.

one point of resistance

Motor starting-current curve for Figs. 262

This

between 1 and

is 2.

to 265.

shown by the decrease in the current curve At the instant of closing the circuit on point

the current increased to 20.77 amperes, also indicated on the vertical at point 2 in the figure. This rise in current caused a 2,

further increase in speed, and the current decreases to 16 amperes, as indicated on the vertical at point 3. When the arm is moved on point 3, the resistance will again be decreased

and the current voltage

ance

is

at,

increased, all the time the speed of, and is rising, until all the resist-

the armature terminal

cut out and full voltage

is

applied to the armature ter-

minal.

Starting Boxes. Figs. 267 and 268 show two types of starting boxes in common use. In the figures R is a resistance

STARTING RHEOSTATS AND CONNECTIONS

251

connected in series with the armature at starting by placing the arm A on contact C. This resistance is gradually cut out

FIG. 267.

FIG. 268.

Four-terminal starting box.

Three-terminal starting box.

by moving the arm across the contacts until it is on the extreme right-hand button. Coil H holds the arm in this position until

ELECTRICAL MACHINERY

252

opened to shut down the motor, or the power fails The connections for series, shunt and comfor some cause. pound motors to the type of starter Fig. 268 was given in

the switch

is

Chapter XVII and will be further discussed in this chapter. is connected in series In this type of box the holding coil with the shunt field, where with the starting rheostat, Fig. 267, the holding coil is in series with a high resistance r, and across

H

the line circuit as in Fig. 269.

FIG.

269.

Shunt motor connected to four-terminal starting box.

Connecting

Up

a Series Motor.

It is

general practice

manufacturers, not only to label carefully the terminals of their apparatus, but to send with each piece of

among

electrical

equipment instructions for connecting up as well as for its care and operation. These instructions are usually accompanied by a well executed picture of the motor, showing how the terminals are located, each terminal being carefully marked and connected to a starting box similar to that shown in Fig. 270,

which represents a

series motor.

However,

as the ter-

minals of various makes of motors

may be brought out differthe connections of one make may of a representation ently, not apply to another, unless certain simple rules are followed.

When

it comes to this, the picture may be dispensed with and a diagram used as in Fig. 271. There is one thing necessary, the diagram must represent the motor it is used for that is, if the motor is series wound, the diagram must represent a series ;

STARTING RHEOSTATS AND CONNECTIONS

253

motor as the one in Fig. 271 if it is shunt wound, it must represent a shunt motor, etc. To connect up a motor from a diagram, first determine which are the armature and which the field terminals, and what kind of a motor it is, either by inspection or by testing. Assume a condition as in Fig. 270, where the two center terminals represent the armature circuit; as can be seen, they come from the brushes and are marked A and A^ The two outside terminals are the field and are marked F and F lm A good practice in connecting up a motor from a diagram is to start from one side of the switch and work around to the opposite side. In this case, start from the positive side of the ;

FIG.

270.

Series motor connected to starting box.

which in Fig. 271 runs to the terminal marked L (sometimes marked "Line") on the starting box. After mak-

switch,

ing this line connection, next in order will come the other terminal on the starting box, which is marked A. By referring to Fig. 271, it will be seen that one wire runs from this terminal to one of the armature terminals marked A, although in this instance, it could be any terminal on the motor as the machine is series connected and there is but one circuit through it.

This

is

true only for the series motor, therefore,

it is

best

make a practice of connecting the machine according to the sketch. The following can be laid down as a hard-and-fast rule

to

for

any motor

:

A (sometimes marked "Arm") on the box to one of the armature terminals only on the starting will motor and this part always be right. Next connect the other terminal of the armature to one of the field terminals, Connect terminal

ELECTRICAL MACHINERY

254

which, in Figs. 270 and 271, are shown A v to F lt although A l could be connected to F. Finally, connect the remaining terminal F on the motor to the negative side of the switch. Now if

the connections are traced out in Figs. 270 and 271,

Fuses

it

will

Snitch

FIG. 271

fuses

6mfeh

FIG. 272

Fuses

Switch

FIG. 273

FIGS. 271 to 273.

Diagrams of series-motor connections.

be seen that they both run the same that is, from the positive terminal on the switch to the L terminal on the starting box; ;

from A on the starting box to the armature connection A; from the armature terminal A^ to the field terminal F 1 and from the remaining field terminal F to the negative side of the switch. In a diagram, such as Fig. 270, the connections may be so

STARTING RHEOSTATS AND CONNECTIONS shown

255

as to produce a definite direction of rotation, while that

in Fig. 271 does not indicate

any particular direction. After the motor has been connected up, however, and is working properly, all that is necessary in order to reverse the direction of rotation,

is to

interchange either the armature or the field

connections, as in Figs. 272 and 273. By tracing out the curIn rent, it will be seen to flow as indicated by the arrows. either case, the connections in Figs. 272 and 273 will give a different direction of rotation from that in Fig. 271.

FIG. 274.

Connecting

Shunt motor connected

Up Shunt

Motors.

to starting box.

What

the series motor can also be applied to

has been applied to In

any type of motor.

is shown a shunt motor connected to a starting box. Every terminal on the machine and starting box is indicated in

Fig. 274

proper location, but, as in the series motor, these connecshown in a diagram, as in Fig. 275. The starting box for the shunt motor, however, has three terminals on it, the its

tions can be

object being to connect the field across the line at the instant of starting. This will be made clear by referring to Fig. 274.

When

the arm on the starting box is moved to the first contact of the starting resistance the field current will flow as indicated

by the arrowheads; that is, direct from the line through the no-voltage release coil on the starting box to field terminal F on the motor, and through the field back to the line, thus applying full-line voltage across the field terminals at starting. This produces a maximum field strength, which develops a

maximum starting torque for a given current in the armature. By tracing out the armature current with the starting-box arm on the

first contact, it will

be seen that

all

the resistance

is

in

ELECTRICAL MACHINERY

256 series

with the armature and the current flows as indicated "by As the arm is moved over toward the no-

the arrowheads.

voltage release coil, the resistance is cut out of the armature circuit and cut in the field. The starting resistance in series

with the so high

field

has

little effect,

for the resistance of the field

compared with the starting

negligible.

resistance, that

This additional terminal

(sometimes marked

"

F

it is

is

almost

on the starting box

Field") does not greatly increase the complications of the connections, although more care should be taken in making the connections than in the series motor. Mistakes in Making Connections. In the series motor the

Fuse*

Switch

Lines

FIG. 275.

Correct diagram of shunt-motor connections.

armature connections on the starting box could be connected to any terminal on the motor and be connected right but not so for the shunt machine, the armature terminal on the starting box must connect to one of the armature terminals on the motor, and the terminal on the starting box marked F (or Field) must always be connected only to one of the field terminals of the motor, as indicated in Figs. 274 and 275, the remaining armature and field terminals being connected together and directly to the line switch. Care should be taken to have the motor connected properly before attempting to start it, as serious results may occur from wrong connections. For instance, if the connections were made as in Fig. 276, the field terminal F on the starting box is connected to the armature terminal A on the motor and the armature terminal A on the starter to the field terminal F on the motor. If the motor is started under these conditions, the no-voltage coil on the starting box will become very hot, and if left in circuit long ;

STARTING RHEOSTATS AND CONNECTIONS enough,

under

it

would be burned

out.

If the

257

motor was starting would be burned

load, the probabilities are that the coil

out almost instantly. Another mistake is that shown in Fig. 277 that is, the field terminal on the starting box is connected to the armature and ;

A,

F,

FIG. 276

A

F,

Swifcf?

-^--,

FIG. 277

FIGS. 276 to 278.

field

FIG. 278 Incorrect diagrams of shunt-motor connections.

on the motor, and the

field

connection on the motor to the

By tracing out the current it will be seen that the fields are across the line, but both armature terminals are connected to the starting box. This will give full voltage across line switch.

the

field,

but zero voltage across the armature, and the motor

ELECTRICAL MACHINERY

258

will not start; although, if the arm of the starting box is thrown over against the no-voltage release coil, it will be held there. No serious results would occur from this connection. Such mistakes are not likely on the large motors, as the field terminals on the starting box and motor are too small to receive the large wires of the armature and line. One of the most common errors in connecting up a motor is that shown in Fig. 278. The armature terminal on the starting box is connected to one of the armature and field terminals on the motor and the remaining armature terminal of the motor is connected direct to -the line. A shunt motor thus con-

nected will start with a good torque at the instant of starting that is, when the arm of the starting box is moved upon the ;

contact the current can flow from the positive side of the line through the field and the armature and back to the opThe posite side of the line and give full field strength. first

1

i

Armature current also flows in the direction indicated, and as :ar as the starting torque is concerned, it will be about the same as that when the motor is connected properly. As soon as the armature begins to turn, it generates a counter-electromotive force. Since the field is in series with the armature the

counter-electromotive force will decrease the effective voltage icross the field and, as the starting resistance is cut out, and

motor speeds up, the field becomes weaker and the motor If the motor is started under load there is danger of the starting resistance being burned out, for instead of the tarting current becoming less as the motor comes up to speed it increases, and by the time two or three points of the resistance are cut out, the starting box is very hot or the fuse has ,he

races.

.

blown.

After the motor has been conmake a test to see that there have been making the connections. This can be done

Checking Connections. nected,

it is

advisable to

no mistakes made in as follows:

After everything is in readiness to start, disconnect the armature connection on the starting box as in Fig. 279, then close the switch and bring the starting-box arm on the first

STARTING RHEOSTATS AND CONNECTIONS motor

259

connected properly, as in the figure, the field circuit will be complete and will be indicated by a spark if the starting-box arm is allowed to drop back to the off posipoint.

If the

is

Care should be taken not to is opened. open the circuit too quickly, as this induces a very high voltage in the field. If the arm on the starting box is brought over tion or the switch

against the no-voltage release coil, it will be held in this position until the switch is open or the arm forced away from the coil.

Sftch + Line

FIG. 279

FIG. 280

FIGS. 279 and 280.

Armature lead disconnected from starting box test

to

motor connections.

With any other combination of connections, the field will be open-circuited if the armature connection on the starting box is open, as in Fig. 280, which is the same connection as in This will be indicated by the absence of a spark Fig. 278. when the starter arm is pulled back from the first contact to the off position, although an open-circuit in the field may be tested for in the same way if the motor is connected up properly.

ELECTRICAL MACHINERY

260

Connecting up a shunt motor from a diagram, as in Fig. 275, involves no particular direction of rotation, but, as in the series machine, after the motor has been connected correctly, and is found to operate properly, the direction of rotation

may field

be reversed by interchanging either the armature or the connections, as in Figs. 281 and 282, respectively. Fig.

281 shows the armature connections interchanged; Fig. 282,

fuses

Switch

+

STARTING RHEOSTATS AND CONNECTIONS

261

duce the same results as explained for that connection. Care should be taken always to remove the jumper between the field and armature connections and to connect it to the opposite This mistake is armature connections with the line wire. often made on machines which have their leads connected to three terminals, as in Fig. 283. Under these conditions, the easiest way to reverse the machine is to interchange the arma-

ture leads at the brushes although either the armature or the field leads may be interchanged at the terminals inside the

frame.

field

POWER.

Shunt motor having only three terminals, connected to

FIG. 283.

start-

ing box.

Connecting

Motor.

Up Compound

The connections

of a

are not necessarily more complicated than series or shunt machine when the compound

compound motor those

of

machine

a is

winding on

To

viewed as only a

series

motor with a shunt-field

it.

consider Figs. 284, 285 'and 286, showing of series-, shunt- and compoundconnections the respectively wound machines. By tracing out the armature circuit in Fig. 286,

illustrate,

which

is

from the positive

side of the switch to the

L

con-

A

nection on the starting box, from the connection on the connection on the box to the motor, and through starting of the line as indiside the armature and field to the negative

A

cated; this

is

found

to be identical

with the series-motor

cir-

Hence, when making the connections of a compound motor, if the armature and series field are connected as a series machine, thev will always be right, and cuit, as

all

shown

in Fig. 284.

that will be left to consider

is

the shunt

field.

If the latter

ELECTRICAL MACHINERY

262

run from box to the shunt field starting motor and from the Sh. F connection to the line this being identical with the field connection in Fig. 285, which bears out the statement that the compound circuit is traced out in Fig. 286,

it

will be seen to

F connection on the (Sh. FJ connection on the

the

;

Fuse6

S*ifch

+

L

A

1 FIG. 284.

Diagram of

series

motor.

Sh.F.

FIG. 285.

Diagram of shunt-motor.

Sh.F,

oh.F

FIG. 286.

Diagram of compound motor.

machine is nothing more nor less than a series machine with a shunt field on it. Care should be taken to determine which is the series and which is the shunt field on the machine before making the connections, for if the series field is connected in shunt the novoltage coil will be burned off the starting box or the fuse

STARTING RHEOSTATS AND CONNECTIONS The

blown.

heavy

lines,

series field is

263

always shown on the diagram in

while the shunt field

is

shown

in light lines.

On

compound motors, except small sizes, the series-field leads are much larger than those of the shunt field. In the small machines, when the shunt- and series-field leads are the same all

the two fields may be determined by testing with a lamp. The lamp will burn brightly through the series field and dimly through the shunt field. After the motor has been connected it is advisable to make a test to determine if the connections size,

have been made correctly.

This can be done in the following

way: Disconnect the armature connection on the starting box, likewise the armature connection on the motor, as shown in

Armature lead disconnected

FIG. 287.

to test connections of

compound-

wound motor. Fig. 287 then close the switch and move the starting arm to the first contact, as shown by the dotted line. If the connections have been made properly, and the switch is opened or ;

the starting-box arm allowed to drop back to the off position, a severe spark will occur. If there are doubts as to which field

has been connected in shunt, do not move the starting-box the first contact, but connect a lamp as shown in

arm upon Fig. 288. will is

If the shunt field

is

connected correctly the lamp will occur when the circuit

burn dimly and a severe spark

open.

If the series field has been connected in place of the

shunt the lamp will burn to full brilliancy and no spark will occur when the circuit is broken. Testing Polarity of Series- and Shunt-field Windings. After the connections have been tested and found to be correct, the next thing to test for is the polarity of the shunt- and

264

ELECTRICAL MACHINERY

series-field coils,

which should be the same.

connect one side of the shunt

field as in

To do

this, dis-

Fig. 289, then start

the motor as a series machine and note the direction of rotation.

Do

shunt

field,

not cut out the starting resistance as tlie motor will reach a dangerous speed. Stop the machine and connect the

then short-circuit the series-field connections as line in Fig. 290 and start the motor again.

shown by the dotted

armature turns in the same direction as before, the and shunt-field windings have the same polarity. If

If the series-

the direction of rotation series field the series-

To remedy

polarity.

is

and

opposite to that produced by the shunt-field coils are of opposite

this, if

the shunt field gives the desired

direction of rotation, interchange the series-field leads. If the series field gives the desired direction of rotation, interchange

the shunt-field connections. shunt-field coils

may

compound machine, then

as a

first

motor,

The polarity of the series- and by taking the speed of the

also be tested

series field as indicated in Fig.

short-circuit the

290 and again take the speed.

If the speed increases with the series field short-circuited, the If the speed decreases, the polarity is polarity is correct.

wrong. Series-

and

shunt-field coils of opposite polarity in a

compound motor

depending upon the strength of the series It may cause a slight increase only in speed from no load to full field. load, but in most cases it will cause one of the following effects: (1) The motor may start and operate satisfactorily under no load. When the load is put on it may suddenly increase in speed and spark badly at the brushes, causing the fuses to blow or throw ihe circuit-breaker.

may

produce several

(2)

effects,

The motor may decrease in speed and if not properly fused may

will blow,

opposite direction; brushes. (3)

this will

The motor may

resistance

is

be accompanied by severe sparking at the

start in

partly cut out

to the point where the fuses stop, reverse and run in the

it will

one direction and when the starting stop, reverse and run in the opposite

direction.

The motor may

fail to start, the armature usually making slight turn in one direction and then in the other If the series-field coils are of the right proportions, a constant speed This condition is seldom, if ever, will be had from no load to full load.

(4)

efforts to

found except

in

motors designed to accomplish

this purpose.

STARTING RHEOSTATS AND CONNECTIONS

265

ELECTRICAL MACHINERY

266

Several of the electrical manufacturers connect one ter-

minal of the

series-

and

shunt-field coils together inside the

machine, after first determining the correct polarity of the field coils. This leaves but five terminals coming to the outside, as in Fig. 291,

the correct polarity

which eliminates the necessity of finding the machine is connected up.

when

To reverse the direccompound motor, it is usually best to

Reversing Direction of Rotation. tion of rotation of a

interchange the armature connections, as in Fig. 292. By comparison with Fig. 286 it will be seen that the direction of the armature current in Fig. 292 is opposite that in Fig. 286, which will cause the armature to turn in the opposite direction.

The direction of rotation may also be reversed by interchanging the field connections where both of the series and shunt terminals have been brought to the outside. Care must be taken to interchange both the series- and shunt-field connections as indicated in Fig. 293. If compared with Fig. 286, it will be seen that the current is in the same direction in both cases in the armatures, but opposite in the field windings. Connecting Up Interpole Machines. What has been said in reference to the simple series, shunt and compound machines also holds true for the same class of machines with Fig. 294 shows the connections of a shunt-interinterpoles.

FIG. 294.

Shunt-interpole motor connected to starting box.

pole machine. The external connections are the same as those in Fig. 274, and the only change in the internal connections is the interpole winding connected in series with the armature. One side of the interpole winding is always connected to one

STARTING RHEOSTATS AND CONNECTIONS

267

side of the armature, with one armature and one interpole connection brought to the outside terminals.

Care should be taken to have each interpole of the proper polarity, which should be the same as the following main pole. This condition

is

shown in Fig.

rotation indicated.

294, for the direction of the machine leaves the

when

Usually, factory the interpole winding is connected for the correct polarity; therefore on most new machines this problem is

already solved, and the connections can be made to the starter and switch the same as for a motor without interpoles.

FIG.

295.

Shunt-interpole motor;

interpole winding disconnected.

After a machine has been taken apart for repairs, unless the connections have been carefully marked, it will be necessary to test the interpoles for polarity when assembling.

A

practical way to do this is, first, to connect the machine in shunt, as in Fig. 295, and see that it is operating properly, with all the starting resistance cut out. The motor may then

be shut

down and one

side of the shunt field opened.

After

carefully marking the 'position of the brushes on the bearing bracket and rocker arm, shift them three or four segments

around the commutator; then connect the interpole winding in series with the armature and start the motor again, this time being careful not to cut out the starting resistance as the motor will race and may reach a dangerous speed. Note the direction of rotation, and if the armature turns in the same direction as that in which the brushes are shifted, the inter-

pole polarity is correct. If it turns in the opposite direction the polarity is wrong. To remedy this, interchange the arma-

ture and interpole connections as shown in Fig. 296 and test the machine again. This time, if everything has been done cor-

ELECTRICAL MACHINERY

268

rectly, the rotation

should be the same as the direction in shifted. After the desired condition

which the brushes were

has been obtained, open the switch and shift the brushes back to their positions and connect up the shunt field to give the

motor the desired direction of rotation.

FIG.

296.

Same

as Fig. 294, but with interpole terminal changed at the armature.

connection

If the interpole is of the wrong polarity, the brushes Avill spark and burn and the machine will have a marked decrease in speed with an increase in load.

A

Fuses

FIG. 297.

Armature and interpole connections changed from those

in

Fig. 294, to give reverse direction of rotation.

Interchanging the shunt-field connections is usually the way to reverse an interpole shunt motor, although the

easiest

direction of rotation can be reversed by changing the terminals of the armature circuit, which must include the inter-

By comparing the direction of and interpoles in Figs. 297 and be opposite, as indicated by the

pole winding as in Fig. 297. the currents in the armature 294. they will be found to arrowheads.

CHAPTER XIX THREE-WIRE SYSTEMS AND HOW THEY MAY BE OBTAINED Voltages for Direct-current Circuits.

Since the two most

extensive uses of electricity are for power decision on what system to use must be based

and

lighting, the

upon

the require-

In industrial direct-current work, the highest voltage at which motors operate is usually between 500 and 600 volts, although in electric-railway work voltages up to 5,000 have been used. Lamps, on the other hand, give the most satisfactory results when designed to operate in the neighborhood of 120 volts. When motors and lamps are to be supplied with power from the same service, it is almost com-

ments of such

services.

pulsory to make the best lamp voltage the controlling element. No matter how desirable it might be to operate the motor load at 500 volts, it would be out of the question to use lamps for such a voltage without operating them in series groups, If the motor least, is very unsatisfactory. were reduced be to 250, it would voltage possible to use lamps on the service, but not advisable. The most satisfactory is method to use lamps which operate at from 110 to 125 volts.

which, to say the

however, the disadvantage of requiring a line wire designed to carry twice as much current as at the higher voltage, which even for the lighting load is a serious handicap, It has,

but which for large power service makes it almost prohibitive. To meet these conflicting conditions a method of distribu-

known as the three- wire system was evolved, which differs from the ordinary circuit, requiring two wires, in the fact that it makes use of three wires as its name The principle implievS. upon which it was developed may be explained as follows

tion

:

269

ELECTRICAL MACHINERY

270 Assume that

required to use one hundred 60-watt 120-volt lamps

it is

amount

to furnish a certain

The

of illumination.

total

would be TP = 100X60 = 6,000 watts.

to be supplied

amount If

of

power

the load were

supplied from a 120-volt two-wire system, as illustrated in Fig. 298,

the current required would be 7 in

drop to be

the line O Tjl

0.06

50

1

W = 6,000

E

120

M were to be restricted to 3

=

#=

=

X0.06 = 150 watts.

ohm, and the power

If

the voltage

volts, its resistance

would have

50 amperes.

lost in

it

would be J 2 # =

(50)

2

Suppose, now, that the lamps were arranged as shown

in Fig. 299; that is, that they were divided into fifty groups of two lamps in series. In that case it would be necessary to impress a voltage of

240 volts across the mains in order to make the lamps burn at their

FIG. 298.

Five lamps connected in parallel on two-wire

The

proper brilliancy.

TJ-T

total load

/>

120 volts.

The

would be the same, and we should

rvri/~\

therefore have I =~=~

= 25

240

hi

circuit.

difference arises

amperes, or half the current required at

from the fact that at 120 volts we have

100 paths through the lamps, each taking 60 watts, or

=0.5 ampere;

whereas at 240 volts we have only 50 paths, each taking that current. If we were to allow the same line loss in the latter case as in the former, I

we should have

(25)

2

X# = 150 watts,

from which we find

Kf\

R=- - = 0.24

ohm, a value four times as great as required before. The voltage drop in the line would be #=/# = 25X0.24 = 6 volts, twice the value at 120 volts; but since the lamps are arranged two in series, a drop of 6 volts across the two is equivalent to a drop of 3 volts across each, the same as on the 120-volt circuit. Consequently, the resistance of the line could be increased fourfold without changing either the loss in it or the voltage drop to the lamps. The cross-section of the wire could therefore be

made one-quarter

of

what

it

was, which means that only one-quarter

of the weight of copper would be required, with a corresponding reducIt may be pointed out here that a tion in the cost of the installation.

THREE-WIRE SYSTEMS reduction to one-quarter of the cross-section diameter.

An

arrangement of lamps

is

271

equivalent to halving the

such, as that indicated in Fig.

299 could be operated satisfactorily if they were always required in pairs, but otherwise the scheme would not be

by turning out one lamp the other would There are additional drawbacks to the series system in that the lamps must be identical, or they will not burn at the same brilliancy, a short-circuit in one results in double normal voltage being impressed upon the other, causing it to burn out, and both lamps are extinguished if either one of them fails.

practicable, since

also be extinguished.

FIG. 299.

Ten lamps connected

series parallel

on two-wire

circuit.

Three-wire System. These undesirable features may be eliminated by the use of a method such as that represented in Fig. 300, which is the typical diagram of a three-wire system. In stead of using a single source of 240 volts, two sources of 120 volts each are connected in series and a connection is made a, and run as a third main wire to the load. commonly referred to as the neutral wire or merely as the neutral of the system, and the other two lines are known as the outside wires. Thus, in Fig. 300, N is the neutral and A and B are the outside wires. It will be noticed that a voltage of 240 volts exists between the outside wires A and B, but that between either outside and the neutral is only 120 volts. Also

at the junction It is

that the neutral is negative with reference to one outside wire and positive with reference to the other.

In the foregoing

it

was shown that in a three-wire system

three sources of voltage are available, two of which have the

ELECTRICAL MACHINERY

272

same value, and the third being equal two.

used

to the

sum

of the other

A

120- to 240-volt system are the approximate values It is, however, almost universally for such service.

entirely feasible to employ exactly the same type of distribution for any other set of voltages, higher or lower, that might

be required in special cases, the 120- to 240-volt system

and whatever follows is

in regard to

equally applicable to them.

To bring out clearly the underlying principle of the threewire system, let us first give our attention to two two-wire systems furnishing current to lamps, as indicated in Fig. 301. Let us assume that we have two generators G and 6r 2 each supplying 120 volts and each furnishing current to twelve ,

FIG. 300.

Ten lamps connected on three-wire

500-watt lamps, making the load on each 12 watts or

6,000

T20~

X

circuit.

500

=

6,000

= 50 amperes when all the lamps are turned on.

The direction of the currents through the mains A, X, Y, and B is that indicated by the arrowheads. For convenience the currents are designated J A) J X) J Y and 1 B respectively. Assume that the two wires X and Y are supplanted by a single wire N as in Fig. 302. The current in this wire will be the sum of the currents in X and Y. Referring to Fig. 301, we find that the current in each

rent in

X

is

from

c

toward

a,

50 amperes, but that the curwhereas that in Y is from & to

is

The resultant current is therefore zero, since one current and the other negative that is, no current flowing inN. Since the current in N is zero, it would appear that the d. is

positive

neutral could be dispensed with.

;

This

is,

however, true only

THREE-WIRE SYSTEMS

50 Amp

IA Ix

b

Y

I

B.

IB

273

=

Y '

50

Amp

50 Amp. 50 Amp.

FIG. 301

MM! A

I

N

IM

B

IB

A ' 50

Amp

=

Amp.

=

50 Amp.

Fia. 302

A

375 Amp

IA

=

Y

-

50

Amp

=

50

Amp

a

X

b

Y

I

B

IB

375 Amp.

Fia. 303

100

A

U=

375 Amp.

N

IN =

12

B

IB

<)<)<)

5 Amp.

-50 Amp.

)O0

Fia. 304

FIGS. 301 to 304.

Shows how three-wire system can be developed from two two -wire systems.

ELECTRICAL MACHINERY

274

when

B

and

is balanced; that is, when there are just as connected between A and N as there are between

the load

many lamps

N or,

to

put

that the current in

it

A

more accurately, when the loads are such has exactly the same value as that in B.

To show what happens when the loads are not the same, let us go back to F g. 301 and assume that three of the lamps between A and X are turned off, leaving only nine in service between these mains as in Fig. 303.

The current

in

A

and

X then

9X500 is

= 37.5

amperes, whereas

Y and B remains at 50 amperes as before. These conditions would also hold if the system were connected as in Fig. 302, and we would have that 7^=37.5 amperes, 7# = 50 amperes and that IN is equal to the difference of the currents Ix and 7y of Fig. 303; namely, 50 37.5 = 12.5 amperes, whose direction will be that of the larger current, that is, If, on the other 7y, and will therefore be from 6 to d as in Fig. 304. hand, the load between N and B is made smaller than that between A and N, we would find Ix to be greater than 7y and the current through AT to be from c to a. that in

The extreme

case is that in which all the lamps on one ;ide are turned off while all those on the other are burning.

Thus,

let

us suppose that the lamps between

FIG. 305.

Lamps and motors

N

and

B

have

connected to three-wire system.

off. Then I B = while I A remains 50 amperes, = 50 amperes. and I N will be equal to That is, when the system is loaded on one side only, the current through the neutral has the same value as that through the outside wire to which the load is connected. This is the maximum value which the current in the neutral can have; as lamps are lighted on the unloaded side the current in the neutral is decreased, until when the loads are balanced no cur-

been turned

7^7^=500

rent flows through

it.

THREE-WIRE SYSTEMS The manner

275

which these currents are then distributed under various conditions of loading, The mains are is illustrated graphically in Figs. 306 to 311. indicated by large pipes which are connected together by smaller pipes L which represent the lamps. The direction of in

in the three conductors,

FIG. 306

FIG. 307

FIG. 308

FIG. 309

FIG. 310

FIGS. 306 to 311.

the currents

Show

FIG. 311

current distribution in three-wire system.

is indicated by the arrows. from the reader are represented by away and those flowing toward the observer by The magnitude of the current is indicated

the circle that ing appears, conductor.

it

Currents flowing the black sections the hatched ones.

by the amount of Where no section-

occupied by the section. means that there is no current flowing in the

is

ELECTRICAL MACHINERY

276

So far we have considered only loads connected between A and N or N and B, that is, across 120 volts. There is, however, no objection to connecting loads directly from A to B that as for example the motors shown in is, across 240 volts

M

Fig. 305. In fact, that is one of the very valuable features of the system. may use the 240-volt supply for the operation of motors or other large loads, and the 120-volt supplies

We

We

for the lamps. thus combine the advantages of a higher voltage for motors with that of a lower one for lamps. The current taken by a load connected to the outside wires has no

whatever on the current in the neutral. Thus, if all the lamps connected to the system of Fig. 305 were turned off and only the motor load remained on the service, the current in the effect

outside lines

A

and

B

would be that required by the load, N would be zero. This is evident, no connection between the motors and the neu-

while that in the neutral since there

is

Should lights be turned on while the motors are in operation, the currents required by them would divide among the three wires exactly as if there were no motor load, and the motor currents in the outside wires would therefore be tral.

increased by the currents required by the lights, while the current in the neutral would be the difference between these lamp currents, which is of course the

same

as the difference

between

M were

20 Thus, where A the and load between H} (namely, amperes, lamp 15 amperes and that between and B (that is, K) were 12 amperes, we would have as the currents in the three lines,

the outside line currents.

if

the motor load

N

N

= 20 + 15 = 35 amperes, I B = 20 + 12 = 32 amperes, and 12 = 3 amperes or, if we choose to use the outside /^ = 15 line current, I N = 35 32 = 3 amperes.

IA

;

In the foregoing the sources of voltage for three-wire systems have been shown as two generators of about 120 volts each, connected in series, with a tap taken off their common point. This method was the one used in the early days of power distribution, but it has been supplanted by two others, which are the ones in common use to-day. One of them employs what is known as a balancer Balancer Sets far Three-wire Systems.

THREE-WIRE SYSTEMS and the other a

set,

special generator

277

known

as a three-wire

generator.

The balancer set consists of two identical machines, which can operate equally well as motors or as generators, and which The simplest arrangement is are directly coupled together. the one shown in Fig. 312, which represents a 115- to 230-volt system. G is a 230-volt generator, and G^G 2 are two 115-volt shunt machines that are alike in all particulars. The gener,

O

<

FIG. 312

FIG. 313

FIGS. 312 and 313.

-Shunt balancer sot on three-wire system.

G may

be far removed from the balancer set G V G 2 which be located near the place where the three-wire service is required. Thus, the 230-volt supply may be received from

ator

,

may

some outside

service,

and the balancer

set installed to

make a

three-wire system available from it. From the diagram it is apparent that the balancer set really consists of two 115-volt

machines connected in

series across

230

volts,

with their com-

N

mon

and their other terpoint n connected to the neutral minals connected to the outside lines A and B.

How Balancer Set Maintains Three-wire Systems Voltage Balanced. The function of the set is to maintain normal voltage

that

is,

115 volts

across each side of the circuit at

ELECTRICAL MACHINERY

278

times and thus provide a true three-wire system. The clear will made of be the contheory operation by comparing all

ditions existing in the machines while the load

with those when

it is

is

balanced,

unbalanced.

Assume that switches SA and SB are closed, but Str left open, as in lamp load is evenly balanced, the voltages from A to N have the same value and will be equal to half the voltage N B Under these conditions there will be no across AB] namely, 115 volts. difference in voltage across the terminals of switch SN, and if it were Fig. 312. If the will to and

to be closed no current would flow in the neutral N.

Next assume that the lamp load

is not evenly balanced while switch will be different open, as in Fig. 313; then the voltage across from that across NB. This can be most readily shown by a numerical example. Thus, if there were one hundred 50-watt 115-volt lamps across

SN

AN

is

AN

and only

The

current of each lamp

= 264

ohms.

fifty across

NB, we would have is

W 50 /==

the following conditions: 115

E R= =

=0.435 ampere.

-

Since there are 100 lamps in multiple across AN, the would be the resistance of one

resistance connected across these mains

264

lamp divided by the across

We

NB would be

100, or

= 2.64

ohms, and similarly the resistance

JLUU

264

=5.28 ohms.

would now have the condition

RI

FIG.

If

N

What I

is

R2

Ohms

Simplified

AN

R

and

2

5?8 Ohms

diagram of Fig. 313.

AB

The voltage across is 230 volts. The question the voltages Ei and and 2 across respectively? the current flowing through RI and R2 we will have Ei=IRi and

between is:

* a&4

314.

which RI that of those

illustrated in Fig. 314, in

represents the resistance of the lamps between

Ez=IR
and B.

AN

E

are

NB

,

To

find /

we

divide the resistance between

voltage across them, which gives /

=

- =29

A

and

B

into the

230 =

2.64 -|-5.28

amperes approxi-

mately. We therefore have #1 = 29X2.64 = 77 volts, and E2 = 29X5.28 = 153 volts. The voltage of Gi is therefore greater than that across AN, whereas the voltage of G? is less than that across NB, as shown in the figure. The difference in both cases is the same; that is, 115 -77 = 38 volts

and 153

115

=38

volts.

This voltage

will exist across the

two

sides

THREE-WIRE SYSTEMS

279

and a voltmeter connected to h and k in Fig. 313 will read 38 volts, consequently current will flow through the switch when it of the switch Str,

is

closed.

Putting this another way,

it

may

be stated that when

as in Fig. 315, machine G 2 is thrown in and in series with them to parallel with the lamps from Then part of the current required by the to N. from

switch

8N

is closed,

N

A

B

.

excess lamps connected from A to N and through the armature of machine

flows along the neutral G 2 causing it to have a ,

armature than G lt consequently The countermachine G 2 increases in speed and with it, 6r a and it its than 6r is made of voltage applied greater x voltage greater current through

its

.

FIG. 315.

Unbalanced load on three-wire system with balancer

set.

becomes a generator supplying a portion of the excess current required by the lamps between A and N. The flow of the current

is indicated by the single arrowheads, Fig. 315. Since both machines must always run at the same speed whatever that speed may be, and since their field currents are

it follows that their generated voltages must be practically the same. The generated voltage of the motor G 2 is its counter-e.m.f The terminal voltage of the generator G i

the same,

.

minus its internal voltage drop due to armature resistance. Likewise the terminal voltage of the motor will be its counter-e.m.f. plus its resistance drop. If the load between N and B is made greater than that from A to N, then machine 6r acts as a motor to drive machine G 2 as will be its generated voltage

a generator. In the foregoing

it

was shown

that,

when a

three- wire

system carried an unbalanced load, the voltage across the more heavily loaded side would be less than the voltage across the

ELECTRICAL MACHINERY

280

more

This effect is considerably more one. the connections are those of Fig. 316, since the decrease or increase in voltage across the lines affects the lightly loaded

accentuated

when

For example, suppose that while no lamps are turned on the voltage of machines G^ and G 2 have been adfield current.

justed by means of their field rheostats R and R 2 until they are both the same, and that more lamps are turned on between

A

and

N

than are turned on between

N

and B.

The voltage

I

Me

FIG. 317

Show methods

FIGS. 316 and 317.

sets

across

NB

AN will

of voltage control of shunt balancer

on three-wire system.

thereby be caused to decrease while that across

the voltage across an decreases and that across nb increases, and consequently the current through will be decreased while that through the field winding 2 1 increases.

That

is,

F

F

will therefore suffer a

The voltage of 6r x further decrease, just as in the case of any self-excited shunt generator, while the counter-voltage of G z which is will be increased. still

,

running as a motor, is increased. The difference in voltage between E and E 2 would of course depend on the difference in the loads on the two sides

THREE-WIRE SYSTEMS

281

It would never be very great even for a conunbalanced condition, but a fluctuation of even a siderably is volts few, objectionable in the case of a lighting load. A very simple way to overcome the previously mentioned difficulty is to connect the field winding of G-^ to the armature terminals of 6r 2 and vice versa as in Fig. 317. In this case a drop

of the system.

,

in voltage across

G

will decrease the current

through

F

2

and

consequently reduce the generated voltage, in this case the counter-electromotive force of <7 2 and an increase in voltage ,

across G^ will have the opposite effect. Similarly, a change in voltage across G 2 will affect the current through F^ and hence

G. Compound-Wound Balancer

the generated voltage of

Another method of two sides of the balancer set is to use compound-wound instead of shunt-wound machines. The connections would then be made as shown in Figs. 318 Sets.

stabilizing the voltages across the

and set

From

319.

when

the load

are

is

the connections in Fig. 318 it is seen that balanced, the two machines of the balancer

operating

as

differentially

connected

compound

motors; that is, the shunt-field ampere-turns oppose those of the series-field winding. When the load is unbalanced, as in Fig. 319, in the machine that acts as a generator (in this case 6rJ the currents in the two field windings have the same direc-

but the motor fields remained in opposition. The latter condition tends to increase the speed of the motor and with it the speed of the generator, which, if the field strength of

tion,

the generator remained constant, would cause the voltage to increase across the terminals of 6?!. However, the series-field assisting the shunt field, therefore as the load is increased the voltage of G will also be increased. Hence the

winding

is

voltage of 6r t will be increased from two sources, one the tendency of the machines to increase in speed and also due to the

generator field increasing in value from the compounding The effect of compounding is effect of the series winding. therefore similar to that found

when

the fields of shunt ma-

chines are interchanged, in that the unbalancing of the load causes the generated voltage of the generator to increase and

ELECTRICAL MACHINERY

282 the

counter-electromotive

force

of

the

motor

to

decrease,

thereby causing the terminal voltages E^ and E 2 to remain practically constant irrespective of the degree of unbalancing. Three-wire Generators. Balancer sets are the most convenient means of providing a three- wire system where a source is available from a separate the unbalanced current required for the sysHowever, there are conditions where it is pref-

of voltage for the outside wires service or

tem

is

when

small.

FlG. 319

FIGS. 318 and 319.

Connections for compound balancer set on threewire system.

erable to use a so-called three-wire generator.

This type of

machine has come to supplant the combination of two generators in series that was originally used in three-wire installations.

The three-wire generator

is

an ordinary direct-current

generator, excepting that in addition to having a commutator its armature is also provided with slip rings that are connected to the

armature winding in the same manner as would be those

of an alternating-current generator. The connections between the windings, commutator and the slip rings and the connections of the

machine

to a three-wire system

may

be represented

THREE-WIRE SYSTEMS

283

The armature winding is represented by A, as in Fig. 320. the commutator by C, and the slip rings by S. The brushes a b on the commutator are connected to the outside wires

A

B

of the three-wire system, and the brushes on the slip are connected to the ends of two equal inductances L rings and L 2 connected in series. The mid-point g of these induct-

and

ances

connected to the neutral

N

of the three-wire system. the system only an alternatingcurrent is supplied to the inductances. The value of this current will be small, similar to the current in the primary is

With a balanced load on

winding of a transformer at no load. When the load is unbalanced the current flowing in the inductances is a combination of a direct and an alternating one, and its value will therefore be changing from instant to instant. Consequently, we shall have to investigate instantaneous values instead of

To study the conditions that exist, refer to Assume that the maximum exciting current taken by the inductances L t and L 2 is 5 amperes and that the inductances have 100 turns each. Then the maximum ampere-

continuous ones.

Figs. 320 to 323.

X

=

200 turns (amperes times turns) is 5 1,000; that is 1,000 ampere-turns are required to set up, through the core, a magnetic field that will generate a counter-voltage in the coils almost equal to the applied volts across the brushes (220 There is only about one or two volts differvolts) Fig. 320. ,

ence between the counter-voltage induced in the inductances and the voltage across the direct-current brushes, when the armature is in the position, Fig. 320, and this difference

between the applied volts and the counter-volts causes a current to flow that magnetized the core. If 1,000 ampere-turns are sufficient to set up a magnetic the core that will cause to be generated a back pres-

field in

sure in coils L! and L 2 almost equal to the applied volts, then if the ampere-turns are increased slightly the counter-pressure

become greater than the applied volts and the inductances will supply power to the system instead of taking t This is just what happens in a from the armature. power three-wire generator, as will be shown in the following:

will

L and L 2

284

ELECTRICAL MACHINERY

THREE-WIRE SYSTEMS

285

320 that the load on the system is unbalanced so that flowing in the neutral. For the position of the armature shown in the figure, up to the point where the unbalanced current is capable of supplying all the ampere-turns, the current in the neutral

Assume

5 amperes

in Fig.

is

through L 2 to the negative point in the armature through the armature winding to the positive brush. Now if the 5 amperes flows through L 2 to the negative brush, it will be in the same direction as the magnetizing current flowing from c to d, consequently will increase the ampere-turns by the value of the current in the neutral times the turns The total ampere-turns required are in L 2 or in this case, 5X100 = 500. 1,000; then, if 500 ampere-turns are supplied by the current in the neutral, flowing through L 2 only 1,000 500 = 500 will have to be supplied by The magnetizing the current flowing into c to d through Li and L 2 will flow

,

,

.

current will equal the ampere-turns divided by the turns in LI and L 2 or 500-^-200 = 2.5 amperes. That is, with 5 amperes flowing in the This neutral and through L2 only 2.5 amperes will flow from c to d. ,

,

will give 2.5

amperes

in Li

and 2.5+5 = 7.5 amperes flowing

in

L

ampere-turns supplied by LI equal 2.5X100 = 250 and those in X 100 = 750 or a total of 250+750 = 1,000, which is correct.

The

2.

L = 7.5 2

current in the neutral was increased to 10 amperes and the through L 2 then this inductance would be supplying 10 X 100 = 1,000 ampere-turns, or the total magnetizing current. Therefore no current will flow from c to d, but the counter-voltage across LI will now be approximately 110 volts, since the magnetic field set up by L2 also If the

total flowed

,

The counter-voltage in L 2 is also approximately through L. 110 volts, and this is just as it should be, since at the instant shown the negative in Fig. 320, 220 volts exists between brush a and point d in the armature winding, 110 of which is absorbed in passes

causing the current to flow through the lamps between A and N, and the remaining 110 volts is used up in causing the neutral current to flow

through L-2 against its ohmic resistance and counter-voltage. Increasing the load 10 amperes between A and over that between and B has caused slight unbalancing of the voltage, so that at the instant in question the voltage across A and B has decreased to a point where it is just equal

N

N

to the counter- voltage in Li| then an increase in ampere-turns in L 2 due to a further unbalancing of the load, will cause the back pressure in and N, and L\ will begin LI to be higher than the voltage between ,

A

to act as

a generator to supply power to the load between

A

and N.

The foregoing is just the condition that we had with the balancer sets, when the load was unbalanced to the point where the machine running as a motor was supplying all the losses in the set, the voltage of the other machine was just equal to the voltage across the side of the system it was con-

ELECTRICAL MACHINERY

286 nected to

with any further unbalancing' of the load one-half

;

was supplied from the machine operating as a generator and the other half from the main generator. Similarly, with the three-wire generator, Fig. 320, if the load is of the current

increased 10 amperes more, making a total of 20 amperes, 5 amperes of this increase will be supplied from the armature

through L 2 from g to d and the other 5 amperes by L^ and will flow from g to c in an opposite direction to the 15 amperes flowing through L 2 as indicated in 500 ampere-turns of Fig. 321. Consequently, the 5 X 100 L t is in opposition to the 15 X 100 1,500 ampere-turns on

and

will pass

will be supplied

=

=

L

which makes the effective ampere-turns equal to 1,500 500 = 1,000, the same as when 5 amperes flowed from c to d through L and L 2J or when 10 amperes flowed through L 2 only. Any further unbalancing of the load will continue to divide up equally between L 1 and L 2 and we will have a current flowing in one direction through L l and in the op2,

l

L

posite direction through 2 Consider the condition that will exist

has

made one-quarter

.

revolution,

when the armature

as in Fig.

position no voltage exists across c d. netizing current will be flowing in L

L and no counterHowwindings L and L 110 volts exists from brush a through L to g, also 110 from a to g through L therefore, the neutral current and

2,

voltage will be generated in the ever,

volts

In this

322.

Consequently no mag.

x

2

;

inductances, 10 am10 through L 2 around through l peres flowing through half of the armature winding to brush a. This is just as it should be, since the currents are flowing through the inductances in opposite directions and are equal; the ampere-turns

will divide

up equally between the two

L and

of one are in opposition to those of the other and no flux is caused to flow through the core, consequently no counter-

voltage will be induced in L^ and L 2 Consider next the condition when the armature has .

one-half revolution, as indicated in Fig. 323. to what we have in Fig. 321 except that L t is the negative point in the armature and

L

2

is

This

is

made

identical

now connected

to

connected to the

THREE-WIRE SYSTEMS

287

Consequently, with the load unbalanced 20 amperes, 15 will flow through L^ to c and 5 amperes through L 2 to d. Now, it will be seen that the current has increased in positive point.

L from 5 in L 2 the t

amperes in Fig. 321 to 15 amperes in Fig. 323, also current has decreased from 15 in Fig. 321 to 5

amperes in Fig. 323. It is this variation in the current that generates the counter-voltage in the inductances as the armature revolves and takes the place of the alternating current that flows in the winding when the load is balanced, or when the machine

is running without load. must be remembered that although the discussion gives the impression of a considerable lapse of time from one position to another, the transition in reality is very rapid. For example, such an armature might be revolving at a speed of 1500 r.p.m., which is equivalent to 25 revolutions per second, and it would therefore make one revolution in 1 / 25 of a second.

It

Since all the changes in current described in the foregoing take place in one-half revolution, they occur in the space of one-fiftieth of a second.

In the explanation of the action of the inductance and how the current divides between them, many other elements that occur to affect the actual conditions have not been considered. This, however, has been done to simplify the reasoning as

much

as possible.

Power Output

Three-Wire Generator.

Suppose that the load is is 100 amperes and the load Y 60 amperes, causing a current of 40 amperes to flow in It will be noticed that while the current leaving the brush the neutral. a is 100 amperes, that entering the 'brush b is only 60 amperes, the difference being accounted for by the 40 amperes that flow from the neutral through the inductances and finally to the armature. The 60 amperes flowing from Y split at the brush 6, half passing through each half of the armature winding to the right and left of the brushes a and 6, as

now unbalanced

of

as in Fig. 324;

that

is,

the load* at

X

indicated.

In the foregoing it was shown how, if the exciting current in the inductances was 5 amperes, with the armature in the position in Fig. 324, then when 10 amperes was flowing in the neutral all the excitation for the inductances would be supplied by one inductance; how, when the armature was in the position shown in the figure, L? would be supplying

ELECTRICAL MACHINERY

288

the ampere-turns, and when the current in the neutral increased above 10 amperes, the increase divided equally between L t and L 2 Therefore, in Fig. 324, 10 amperes of the 40 in the neutral will flow through L 2 to supall

.

ply the ampere-turns and the remaining 30 will divide 15 through through L 2 making a total of 15 amperes in LI and 25 amperes in ,

L and 15 L The r

2.

25 amperes flowing through L 2 enters the armature winding at d and divides 12.5 amperes flowing through the winding to the right of brushes b and a and 12.5 amperes to the left, making a total of 30 + 12.5=42.5 amperes flowing in each half of the armature, or a total of 42.5+52.4 = 85 amperes. This, added to the 15 amperes delivered by LI, makes a total of 85 15 = 100 amperes, leaving the positive brush as indicated.

+

FIG.

324.

Shows

distribution

of current in three-wire

generator.

The next thing is to investigate the power distribution in the system. The entire load on the generator is the sum of the loads X and Y and the excitation required by the inductances LI and L 2 .

X

100 amperes and the voltage across it is 110 = 100X110 = 11,000 watts. Similarly, Y = 60X110 volts, we have = 6,600 watts, and the watts taken by LI and L 2 at the instant shown in the 5X220 or 10X110 = 1,100, making the total load that the generator has to supply equal to 11,000+6,600+1,100 = 18,700 watts. The total Since the current in

is

X

current -flowing in the armature

is

42.5+42.5=85 amperes;

this value

times the voltage gives the load supplied by the armature, equals 85X220 = 18,700 watts, which checks with the sum of the component loads on the external circuit.

In looking at Fig. 324, at first thought it might appear that the 15 amperes through LI should be considered in the current supplied by the armature; the fact is, however, that this current is not supplied from the armature, but is 220-volt power supplied to the system by the

armature and

is

transformed into 110-volt power through the medium This will be mo'-e readily understood by referring

of the inductances.

THREE-WIRE SYSTEMS

289

At the instant shown in Fig. 324, if the inductances were disconnected from the brushes on the slip rings and connected across the line, as in Fig. 325, the conditions would be identically alike except that the current flowing through LI would not pass through brush a. The current in the armature (60 amperes from load Y and 25 amperes to Fig. 325.

through load

L

2 ),

that

is,

7=/j3-f/d = 60-|-25

i

FIG. 325.

Same

= 85

amperes.

This gives

-as

as Fig. 324, but with the inductance connected to the external circuit.

the total power supplied by the generator equaling "7=220X85 = 18,700 watts, which check up against the lamp load of 17,600 watts, plus the

power required by L\ and or

10X110 = 1,100 watts

L

2

at this instant, to excite them, of

or a total of 17,600

+ 1,100 = 18,700.

5X220

CHAPTER XX DIRECT-CURRENT GENERATORS IN PARALLEL When a with electrical power, a single

Reasons for Operating Generators in Parallel. constant load

is

to be supplied

be installed to provide the energy, its capacity being just large enough to take care of the load, since On the other it operates at highest efficiency fully loaded. when one the varies from load part of the day hand, greatly

generating unit

to another,

it

may

would be uneconomical

to

supply power from a

single unit because it would be operating considerably below full load, and therefore at a low efficiency, for much of the

usual to provide two or more units of such capacities as will allow of their operation time.

Under such circumstances

it is

at as nearly full-load conditions as possible at all times. When two or more generators are furnishing power to a

common load, they are said to be operating in parallel. In order that two or more generators operate together satisfactorily each

must take

its

due share of the

load.

The ques-

is, therefore, Will any two generators operate successfully when connected in multiple? In order to answer this it is

tion

necessary to inquire into the factors controlling their behavior. The current delivered by a generator is equal to the difference between the generated and terminal voltages divided by the armature resistance. That is, if / represents the armature current, E g the generated voltage, E the ter-

minal voltage, and

E

ff

-E R

,

from which

R

the armature resistance,

it is

relations to each other.

seen that If the

E g and R

we have

/

=

must bear certain

machines are of the same

size,

the relation between the values of these quantities in each 290

DIRECT-CURRENT GENERATORS IN PARALLEL

291

must be such that 7 is the same in all if they are of different sizes, the relations must be such that / in each case is propor;

tional to the capacity of the machine.

on Parallel Operabetween current, armature generated voltage and terminal voltage are gov-

Effects of Generator Characteristics tion.

The

resistance,

relations

existing

erned by the characteristics of the machine. Their effect upon its behavior is most readily determined by reference to the

50

40

40

30

20

Load Current FIG.

326.

60

50

70

in

External characteristic curves of two shiut generators of unlike design.

.

external characteristic curve of the machine.

This, as will be

a curve showing the relation between the current delivered by a generator and its corresponding terminal

remembered,

is

For a shunt generator its shape would be such as shown in Fig. 326, which represent the characteristics of generators of the same size but of different designs. Although having the same voltage at no load

voltage.

that of the curves

namely, 120 volts

they do not undergo a like decrease in For example, at 100

voltage for similar increases in current. volts the

machine of curve

A

would be taking 60 amperes,

ELECTRICAL MACHINERY

292

B would be taking only 40 amperes. increasing the excitation of B or decreasing that of A, the machines could be adjusted to take the same current at 100 volts, being 60 amperes in the former case and 40 amperes in whereas that of curve

By

the latter. If both

machines had precisely the same characteristic, the

current in each would be half the total at

all times, since at

the same terminal voltage the same value of current would be delivered by each. Consequently, it is concluded that gen-

10

10

W

30

Load Current FIG. 327.

50 tn

External characteristic curves of two compound generators.

erators having unlike external characteristics will not operate in parallel satisfactorily and, conversely, that such as have If the generators are of similar characteristics will do so.

must be such that the among them in the propor-

different capacities, their characteristics total load will

always be divided

tion of their ratings.

The same course

of reasoning can be

applied to any other type of generator, such as the series and compound. Discussion of the former may be omitted, since

H

finds

no practical application, but the

latter is of great

DIRECT-CURRENT GENERATORS IN PARALLEL importance. The compound generator of the direct-current machines.

The external

characteristics of

is

293

the most important

overcompounded generators

have shapes similar to those of the curves illustrated in Fig. 227, in which F is the characteristic of a machine more heavily compounded than the one whose characteristic is G, but of the

same capacity. The dissimilarity in the curves will have the same effect as in the case of shunt generators that is, it will prevent the machines from assuming equal shares of the load. As with shunt generators, they could be made to assume the. same terminal voltage for like currents by manipulation of the shunt-field rheostats, but they would not make such a division :

A

FIG. 328.

Simplified diagram of two

compound-wound generators

con-

nected in parallel.

However, if possessed of identical external characteristics, they will do so, in which respect also they behave as shunt machines would. It would therefore be of the load of themselves.

concluded, as in the case of the shunt type, that machines with unlike characteristics will not operate well in parallel, whereas those of like characteristics will.

The conclusions regarding shunt generators in parallel are borne out in practice, but it is found that compound generators refuse to operate properly even when their characteristics

are identical, unless an additional feature,

equalizing connection,

is

known

as the

provided.

Compound Generators

in Parallel.

To

facilitate the dis-

cussion of the points involved reference will be

made

to Fig.

328, which shows two compound generators, 1 and 2, connected to the mains A and B, supplying current to a load Zr*

ELECTRICAL MACHINERY

294

A

to be the positive and B the negative; then the Suppose direction of the currents through the shunt and series fields

in the armatures will be that indicated by the arrowheads. Further suppose that the generators are of the same capacity and design, and that they are overcompounded alike. Also that it has been possible to place them in parallel and that

and

each

is

load L.

therefore delivering half the current supplied to the The following will show that the machines cannot

maintain this assumed condition.

Anything

affecting the generated voltage of either 1?

will affect its load current, since I

machine

T?

g

K

For example,

the speed of generator 1 might increase slightly, thereby g and a consequent increase of /. causing an increase of

E

Since the current flows through the series field of the machine, the increase in its value will cause an increase in the seriesfield excitation

shunt-

and

and a corresponding increase in the combined The generated voltage will

series-field strength.

therefore be further increased, causing a yet greater current to flow with the effect of again increasing the field strength

and in turn the current.

It will be seen that the action is

cumulative; that is, that, once started, it will continue unchecked. Just how far it will proceed depends upon the

amount

When

of overcompounding.

the current delivered by

No. 1 increases, that of No. 2 must decrease, since the load L remains practically unchanged. The effect of a decrease in current of No. 2 is exactly the reverse of what an increase would have as described in the case of No. 1 that is, the initial decrease would cause a diminution of excitation which would further reduce the current delivered, again affecting ;

the field strength. This action will continue until No. 1 carries the entire load and No. 2 carries none. Generally, the action does not cease here, but proceeds until No. 2 is running as a motor driven

from No.

1

;

that

is, it

line instead of delivering its share to

takes current from the

it.

The

direction of the

current through its armature would consequently be reversed and therefore that through its series field.

DIRECT-CURRENT GENERATORS IN PARALLEL The

295

shunt-field current would, however, continue to flow

same direction as before, since the field remains connected to the mains as before. The two fields would thus be in the

opposed to each other, with the shunt

field

predominating at

the time the series field begins to reverse. The total field strength will of course be less than that due to the shunt field alone,

which

will tend to

make

the machine speed

up

since it

running is, however, connected to its prime mover whose speed generally cannot be rapidly accelerated, as a motor.

is

and the motor as before even is

It

therefore compelled to run at the same speed though its field be weaker. The counter-e.m.f.

is

therefore reduced, which increases the current.

This

in-

crease in current increases the series-field current, thereby causing a still weaker field and a consequent further increase in

current.

The action would continue until the current

became so great that the safety devices protecting the machine from excessive current would operate. In practice it is found that the sequence of events enumerated in the foregoing takes place in a very short period of time; in fact, the action gives the impression of being an instantaneous one. No sooner has the incoming machine been

connected to the mains than the circuit-breakers open. Often the instantaneous rush of current is so rapid that before the circuit-breakers can interrupt its flow it has become so great its effect in flowing through the series field is to cause the

that

series-field excitation to

exceed that of the shunt

field

and

consequently reverse the direction of the magnetic field. It would then be found that, when the machine was again

would build up in the opposite direction to When such a condition exists the previously had. Here polarity can be corrected as indicated in Fig. 329. machine A is assumed to be the one with reversed polarity. started, its voltage

that

it

First open the armature circuit by lifting the brushes, or still by opening the armature and equalizer connections, as in the figure, then close the switch to the bus bars for a

better

short period, which will cause a current to flow through the shunt-field winding in the correct direction to give proper

ELECTRICAL MACHINERY

296

In Fig. 329

polarity.

it

will be seen that the current

the shunt-field windings of both machines

is

through

the same direc-

consequently they will both have the same polarity.

tion,

FIG.

Two compound

329.

Equalizer equalizer

machines.

is

generators connected in parallel.

Connection. As indicated in Fig. 328, the a connection between the terminals ab of the

With switch

factorily, but with

it

s

open they would not operate

closed the equalizing connection

satis-

would

and the machines would then operate together The successfully. equalizer connection must always be between the junction points of the armature and series field. That is, it could not be made between c and d, for example, nor c and b, nor a and d. Moreover, it must be of a very low resistance. To more clearly understand the principle involved, reference >e

established

made to Fig. 330, which duplicates the connections of Fig. 328, but shows a slightly different location of the series

will be

fields.

the load is the same on each generator in Fig. 330 and the serieswindings F and F2 have equal resistance, the current from the negative bus will divide equally through the series windings back to armatures 1 and 2. Assume that one of the machines takes more than If

field

v

its share of the load, say 80 amperes on machine No. 1 and 60 amperes on machine No. 2, or a total of 60+80 = 140 amperes. If the series-field windings have equal resistance then the current will continue to divide equally between them, and each will take 70 amperes, but at the equalizer

80 amperes will flow to armature No. 1 and 60 amperes to armature No. 2. This creates a condition where armature No. 1 supplies 80 but only has 70 amperes in its series windings, where No. 2 armature supplies

DIRECT-CURRENT GENERATORS IN PARALLEL

297

only 60 amperes, but has 70 amperes through its series winding. On account of the excess current in the series field winding of No. 2 machine the voltage tends to rise and cause the machine to take more load. On machine No. 1 the current in the series winding is less than that required to compensate for the volts drop in the armature, consequently, the voltage of this machine will tend to decrease and give up part of its load to machine No. 2, thus restoring an equal division of load between the

two machines.

If

machine No. 2 takes an excess load the reverse action

of that described in the foregoing, takes place 1,

FIG. 330.

>.

A

.

and the load

is

maintained

T

Modification of diagram, Fig. 328.

Any desired division of the load can be obtained by proper adjustment of the series winding, both as to ampere-turns and resistance. The machine to take the greatest load must have the lowest resistance in its series-field windings. equally between the two machines.

Since in a shunt generator the voltage falls

off as

the load

increases, two or more machines will operate in parallel satisIf one machine takes an excess of the load this factorily.

tends to lower the voltage on this machine, and as the other machine has been relieved of some load its voltage tends to increase. This causes the lightly loaded machine to take more load and restore the balance between the two machines.

There are a number of Putting- Generators into Service. points that require attention when cutting-in a generator to operate in parallel with others and also when cutting-out one In the main the procedure for shunt and compound generators is the same, but there is some difference, and the two will therefore be treated separately. In Fig. 331 are shown the connections of a self-excited shunt generator G that is part of a system of other generators that

is

so operating.

ELECTRICAL MACHINERY

298

connected to the busbars

M

.

Within the outline

are repre-

sented the devices on the switchboard used in controlling it. They consist of the main switch 8, which is, of course, open

when the machine

is

not in service, the field rheostat

R

for

adjusting the current through the field coils F, a voltmeter V, connected to a switch T for obtaining the voltage at the bus-

M

an ammeter A for measuring the current output of the machine; and circuitbreakers CC to protect it against overload. We will assume that while the set is at rest it becomes necessary to place it in bars

service,

FIG. 331.

and the generator

respectively,

and follow the steps necessary

to accomplish this.

Connections for paralleling a shunt generator with one or more machines.

The first thing to do is to start the unit and bring it up to normal speed, after which switch T is thrown to position b and the field current is adjusted by means of the rheostat R until the voltmeter V indicates approximately normal voltage. The switch T is then thrown to position a and the reading of voltmeter V is noted, upon which T is thrown back to & and the voltage of the generator adjusted to exactly the same Switch T is value by further manipulation of rheostat R. then thrown from "b to a and back again to assure the operator that the two voltages are of like values, and if they are not quite so the rheostat R is manipulated until they have been

DIRECT-CURRENT GENERATORS IN PARALLEL The switch 8 equal. the generator to the mains

made

then closed, thereby connecting

is

M

299

If the voltage of the generator has been carefully adjusted to the same value as that of the mains, the ammeter A will indicate no current. .

If the voltage of the generator was a little lower than that of the mains, the ammeter will show a reversed deflection, showing that current is being taken from the mains by the

machine and if the generator voltage was somewhat greater than that of the mains, the ammeter would indicate in the correct direction and its reading would be the current deliv;

ered to the system by the machine. Having connected the generator to the mains, it is then made to assume its share of the load by adjustment of resistance in the rheostat R. The resistance of the rheostat is reduced until the ammeter

A

indicates that the generator of it.

When

it is

is

delivering the current required

desired to remove a generator from the system, is increased until the ammeter

the resistance of the rheostat

R

A

reads approximately zero current, showing that the voltage generated by the machine is exactly equal to the voltage of the mains.

The switch 8

generator, whereupon unit is dead.

its

then opened, disconnecting the prime mover is shut down and the is

In Fig. 332 are shown the connections for a compound generator corresponding to those for a shunt generator in Fig. 331, except in addition are shown the series-field winding

F

a

to

and the equalizing connection B, which* is a busbar similar M^M 2 The operation of placing the machine in service, .

paralleling it with another, is identical with that for a shunt generator, except that its voltage is adjusted to be a few volts less than that of the mains instead of equal to it.

that

is

The reason for

this

is,

that

when

the switch

8

is

closed current

F

immediately flows through the series-field winding Sj causing the generator to be supplied with excitation additional to that

and consequently inducing a greater generated voltage. Just how much less than the line voltage the open-circuit of the shunt field

ELECTRICAL MACHINERY

300

voltage of the incoming generator should be is a matter of Having once been determined, the operator adjusts it to that value whenever cutting-in the machine. method of trial.

A

FIG. 332.

Connections for paralleling a compound generator with one or more machines using a three-pole switch.

FIG. 333.

Connections for paralleling a compound generator with one single-pole and a two-pole switch.

or

more machines, using a

avoiding the complications indicated in the foregoing is to substitute a double-pole and a single-pole switch for the threepole switch S of Fig. 332. Such an arrangement is shown in In this case the generator is adjusted to approxiFig. 333.

DIRECT-CURRENT GENERATORS IN PARALLEL mately line voltage with the

field rheostat

301

and then the double

pole switch S d is closed, thus connecting the series field F 8 which are already connected the series across 2 and B, across Current fields of all the other generators serving the busbars.

M

consequently flows through F s as explained in the preceding paragraph, thereby increasing the voltage of the generator. This is now adjusted to exactly the same value as the line voltage, as in the case of shunt generators,

and the

single-pole

S is then closed, thus placing the machine in service. Having connected a compound generator to the line, the load upon it is adjusted in the same manner as for a shunt

switch

s

generator; namely, by increasing its shunt-field current by cutting resistance out of the shunt-field rheostat. When it is desired to take

it

out of service, the shunt-field resistance

is

increased until the excitation has been reduced sufficiently to

cause the ammeter

A

of Fig. 332 to indicate zero current, of Fig. 332 or switches Ss and S d .of upon which switch 8 Fig. 333, as the case may be, are opened and the set shut down.

In the case Parallel Operation of Three-wire Generators. of a compound three-wire generator there is one complication When the system is balto which attention must be drawn.

anced there will be no current in the neutral N, Fig. 334, and

FIG. 334.

Three-wire compound generator; series-field winding in one section.

the current in

B

will be the

same as that through A, and the

conditions will be the same as in a two-wire system. Assume that the load is unbalanced for example, that more lamps are ;

ELECTRICAL MACHINERY

302

X

than at Y. There will then be less current in turned on at than in A, and consequently less current through the series field F than is delivered by the armature G. The extreme case turned on and all at Y turned off. would be all the lamps at

B

X

There would then be no current through B and therefore none through F. That is, although the generator would be delivering half its rated output, there would be no current through its series field and consequently no series-field magnetic flux to keep the voltage from dropping with increase of load. On the other hand, if all the lamps at Y were to be turned on and those at turned off, there would be no current in A,

X

and

B

to flow

would carry

full-load current, causing full-load current

The series-field excitation would therefore and Y would be if all the lamps in both and it would give as much compounding in

through F.

be the same as

it

X

were turned on, the one case as in the other. To overcome these disadvantages it is necessary to connect one half of the series-field winding into one side of the circuit

and the other half into the other side, as F^ and F 2 335. This offers no difficulty, since there are always

FIG. 335.

Three-wire compound generator into

two

poles,

and

two

;

series-field

in Fig. at least

winding divided

sections.

since each pole has its share of the series-field

winding wound upon it. With this arrangement the currents in A and B each produce the correct amount of magnetization. If the machine is a multipolar one it would be necessary

DIRECT-CURRENT GENERATORS IN PARALLEL

303

somewhat for the connections which the as shown in Fig. 336, represents three-wire series-field windings of a six-pole compound-wound (The shunt-field windings are not shown.) The generator. discussion in reference to series windings applies equally to those on interpoles. That is, when a three-wire generator is of the interpole type, half the interpole windings must be connected into one side of the system and half into the other,

to connect alternate series-field windings in series

FIG. 336.

Series winding on three-wire generator.

when too little current would them and others when there would be too much.

otherwise there would be times flow through

The distribution of the windings can be accomplished in the same manner as with the series windings namely, by connecting alternate ones in series and then connecting one group to the positive armature terminal and the other to the negative ;

one.

If the machine is a compound one, each group of interpole windings would of course be connected in series with one group of series-field windings. In connecting up the series

and interpole windings of such machines, great care must be

ELECTRICAL MACHINERY

304 exercised to

make

the connections in such

manner

that the

magnetic poles created by the current through them shall have the correct polarity. If the poles created by the shunt-field and 8 in Fig. 336, the conwinding are those indicated by

N

nections for an interpole machine would have to be those of Fig. 337 if the direction of rotation is that indicated by the If the direction of rotation were the reverse the connections would have to be changed to give correct polarity.

arrow K.

FIG.

337.

When

Series

and interpole winding on three -wire generator.

dealing with three-wire systems of large

demand

or

having heavy peak loads at certain hours of the day, the same factors govern as in any other electrical system, and it may become necessary or advisable to install several units to supply the load instead of a single one. When such is the case, three-wire

generators are operated in parallel.

A

diagrammatic representation of two compound-wound three-wire generators connected in parallel is given in Fig. 338. As explained in the foregoing, the series-field windings three-wire generators must be divided into two parts, one of which is inserted between one terminal of the of

compound

DIRECT-CURRENT GENERATORS IN PARALLEL

305

armature and a main and the other between the other terminal and main, as at F a} F b F c and Fd With compound generators in parallel it was shown that it is necessary to have an equalizing connection joining those terminals of the armatures to which the series-field windings are connected. In the .

,

case of the three-wire generator

it

is

therefore necessary to

provide two equalizing connections, since both the armature terminals in this case have series-field connections made to them. These two equalizers are shown at Z and Z 2 and connect a to c and & to d. The series fields Fa and F c are connected between Z^ and the main A, and the series fields F b and F d are connected between Z 2 and the main B. The midpoints g l and g 2 of the inductances L a L b and Lc L d) are con,

.

nected to the neutral main N. J*.

FIG.

338.

Schematic diagram of two three-wire generators connected in parallel.

The two equalizers function in exactly the same manner as does the single equalizer used when two-wire compound generators are operated in multiple. Thus, Z x connects a and

F

F A

in multiple

and thereby causes whatever current

to split evenly

and

Fd

between them.

in multiple

Likewise

and causes the current

Z

2

flows in

connects

B

in

F

l

to divide

equally between them. In order that such shall be the case, of course necessary that the resistance of*F a equal that

it is

of

F

c

and the resistance

of

F

shall equal

Fd

,

as

was

ex-

compound generators were discussed. Although only two generators are shown in the diagram, it

plained at the time that

ELECTRICAL MACHINERY

306 is

to be understood that

any number of machines may be con-

nected in parallel in the same manner.

The method

of operating three-wire

compound generators multiple essentially the same as that outlined for ordinary two- wire machines. The chief points of difference

in

is

are that with the three-wire generator

we have

the inductances

to connect to the alternating-current side of the

machine and

we must make

provision for connecting the mid-point of the inductances to the neutral main at the same time that con-

that

nection

is

made between

the direct-current side of the machine

and the outside mains. It is also necessary to provide two ammeters for each machine since the current in the two outside mains is not the same when there is current flowing through the neutral. It is also desirable to have a third ammeter to measure the cur-

rent in the neutral.

If one is installed, it should be of the

type that has its zero point at the middle of the scale and reads in both directions from the center, since the direction of the current in the neutral may be in either direction, depending upon which side of the system is the more heavily or lightly loaded. The ammeters that are used to measure the currents in the outside mains should be connected between the series

and the armature. Thus, in Fig. 338 the ammeter should be connected between a and a lt & and 5 X etc. They will then field

,

measure the current actually delivered by the armature. If the ammeter were to be. connected between the series fields and the mains, as for example, between x and y, the current measured would be that through the series field. Since the currents through all the fields between A and Z^ or between B and Z 2 are the same, the reading of the ammeter

would be no indication of the load carried by the machine. This condition also introduces a complication when interpoles compound generators. Since the current through

are used on

the interpole windings must change with the armature current in order to secure good commutation, it follows that the inter-

pole windings must be connected into the same part of the circuit as are the ammeters.

j

CHAPTER XXI

FUNDAMENTAL PRINCIPLES OP ALTERNATING CURRENT Dynamo-electric Machines Fundamentally Generate Alternating Current. Before taking up the studies of alternating current the student should get firmly in mind the

down in Chapter X. In this chapter it is dynamo-electric machines are fundamentally alternating-current machines, and may be converted into principles laid

shown that

all

direct-current generators by the use of a commutator.

Figs.

X

146 to 149 in Chapter explains how a coil with its ends connected to two rings will deliver an alternating current to an external circuit. This current, it is explained, may be represented by a curve, Fig. 152. For a thorough understanding of alternating current one must first get a clear conception of the underlying principles,

which

will here be taken

up

briefly.

An angle may be considered as generated by the radius of a circle revolving about a point at the center of the circle; thus, in Fig. 339, if the radius OB be revolved counter-clockwise about the point from the horizontal position OA to the position OB, angle BOA will be generated, and if a perpendicular is let fall from B to the opposite side OA, In any right triangle the sine the right triangle will be formed. of either acute angle is the ratio of the opposite side to the hypotenuse;

OBD

BD thus, in Fig. 339, the sine of the angle

(3

=

OD -.

OB

If side

When OB

OB

<

equals

OB

,

or the sine of angle

BD is

given a value of unity, sine

<

will equal

=BD.

1

coincides with

equal zero;

OA, angle



will

equal zero and

therefore the sine of angle $ will be

has revolved up into the vertical position OE,

307

BD

OB will

BD

= 0.

will also

When OB

be equal to and

ELECTRICAL MACHINERY

308 coincide with

OB;

therefore, sine 90 deg. will be

OB =-1 =

BD

1.

Thus

it

1

seen that the sines of angles between zero and 90 deg. vary from zero to unity; these values are given in Table II. The cosine of an angle is the ratio of the adjacent side to the hypotenuse; is

thus the cosine of an angle 0=777;.

OB

with and

As

OB

until,

OD

equal to the side

is

OD;

When

angle

<

is

therefore, cosine

<

OB

zero,

deg.

coincides

OD = -1 =

=

OB

1.

1

revolves around to the vertical position OE, OD decreases in length OB coincides with OE forming an angle of 90 deg. with OA,

when

becomes

therefore,

zero;

cosine 90 deg.

=

OD = OB

=0.

From

this

1

it will be seen that the cosine of the angle decreases from unity for an deg. to zero for 90 deg., which is just opposite to the sine. angle of The values given in Table II, for the sines and cosines are the natural functions and should not be confused with the logarithmic values.

FIG. 340

FIG. 339

FIGS. 339 to 341.

FIG. 341

Illustrating harmonic motion.

Distance Traveled and Rate of Motion.

One must be

careful to distinguish between the distance through which a train going from New body moves and its rate of motion.

A

York to Philadelphia travels approximately 90 miles, and if the run is made in 120 minutes, the average rate of motion would be

% mile per minute. In this case 90 miles the dis% mile per minute the average rate of is

tance traveled and

minute that the train is in motion it uniform rate and may have uniform motion that is, it will move through

If for each

motion. travels

% mile,

be said to

it

will be traveling at a ;

equal units of space in equal units of time. For this to be mile per possible the train would have to be moving at

%

minute at the beginning and at the end of the run. If it were to start from rest and travel the 90 miles and come to rest again in 120 minutes, although the average rate of motion

PRINCIPLES OF ALTERNATING CURRENTS

309

% mile per minute,

the train would have a varying cannot start and stop instantaneously, but must start from rest and gradually increase to full speed, which mile per minute to make up for lost must be greater than

would be

motion, for

it

%

time at starting and stopping and gradually coming to rest at the end of the run. The rate of increase in motion is called acceleration, whereas the rate of decrease

is

called decelera-

There are several kinds of varying tion, is most important in the study of that but one the motion, current is harmonic motion. alternating In Fig. 340 suppose a body to be moving around the semicircle ABC at a uniform rate and also a second body moving across the diameter AC at such a rate that it will always remain in the same vertical plane as the body moving around the path ABC; that is, when the body moving around the semicircle is at a the body moving along the diameter will be or retardation.

at

a',

as indicated, or

will be at

when

From

the

first

body

is

at b the second

be seen that although the body moving around the semicircle has a uniform motion, that moving across the diameter has a varying motion start&', etc.

this

it

will

ing from rest at A, reaching a maximum at 0, and then decreasing in speed until it comes to rest again at C. This body is

said to have harmonic motion

and

will be referred to later.

When

the two bodies are in the vertical plane BO, Fig. 340, they are both moving in the same direction and at the same rate. If the radius of the circle in Fig. 341 is let

ABC

to represent the

maximum

rate of motion of the

body moving and consequently the rate of motion of the body moving around the semicircle ABC, the rate of motion of the former at any point along the diameter is equal to the perpendicular distance (such as DN) between the two along the diameter

AC

any instant. It will be observed that DN increases from zero at A until it is equal to the radius BO at B.

bodies at in value

Following

is

a proof of this rule:

The direction of motion

at

any instant of a body moving

in a circle is always tangent to the circumference, or at right angles to the radius. In Fig. 341 consider the instant when

ELECTRICAL MACHINERY

310 the body

is

to

which time

at D, at

angles to the radius

DO.

Draw

it

will be

the line

moving

DF

at right

at right angles

DF represents the direction of motion at this On DF lay off the distance DE equal to the radius

then

DO;

instant.

DO

(equal to the rate of motion of the body moving around the circle). Draw and from parallel to the diameter E let fall a perpendicular to DH, thus forming the right tri-

DH

angle

DEG. Now

around the

AC

DE equals the rate of motion of the body DG will equal the rate of motion at this

if

circle,

moving along the diameter AC. The right and DON are similar (their corresponding

instant of the body triangles

DEG

sides being perpendicular)

But

as

DE

DN.

;

was made equal

therefore to

OD,

it

DE OD = GD :

follows that

GD

:

DN.

must

DG

equals the rate of motion of the body the moving along diameter, DN, the perpendicular distance between the two bodies at any instant will also equal the rate

equal

of

Since

motion of the body along the diameter.

FlG. 343

FlG. 342

FIGS. 342 and 343.

To

Relation between constant and harmonic motion.

342 that the body traveling around the rate of 15 ft. per sec., which is also the maximum rate along the diameter AC. Let the radius OA represent this rate, and consider the instant when the body has moved through 30 deg. on the semicircle, as shown by the position D. The perpendicular DE will represent the rate along the diameter AC at this instant. By illustrate,

semicircle

joining

ABC

and

assume

is

in Fig.

moving at the

D the triangle ODE will be formed.

whence DE = OD

Then

sine 30 deg.

=

T)T? ,

OD = 15 ft. per sec. and DE = 15X0.5 = 7.5 ft. per

In this case sine 30 deg. 30 deg. =0.5 (see Table II). Therefore, sec., which is the rate at which the body along the diameter is moving This must not be confused with the distance traveled, at this instant. as the body has moved only the distance AE, which in this case is about 2.1 ft., but in moving this distance the body has been accelerated to about sine

j

I

j

>

PRINCIPLES OF ALTERNATING CURRENTS

311

= 15X0.866 = 12.99 ft. = per sec. At 60 deg. D'E' OD' sine 60 deg. AE' or 7.5 ft. In each distance the moved per sec., and the body has is proportional case the rate of the body moving along the diameter to the sine of the angle through which the body traveling around the

7.5

ft.

AC

On the line A'O', Fig. 343, lay. off distances proporcircle has moved. in Fig. 342, tional to the divisions of the circumference of the circle and horizontally the various points on the circle to meet the

ABC

project

vertical ordinates erected at corresponding points

on the

Line A'O',

thus

By joining these points a respectively. locating points A', b, c, curve will be formed as shown. The vertical distance from any point etc.,

curve to the base line AO will be proportional to the sine of the corresponding angle in the circle ABC. For this reason the curve thus formed is called a sine curve. This is the curve of electromotive force generated by a conductor revolving about an axis at a constant speed

in this

in

a uniform magnetic

field,

as will be

shown

later.

Dynamo-electric Machine. When a conductor is moved in a magnetic field so as to cut the lines of force there is always a difference of potential developed between its terminals, or in other words an electromotive force This is the fundamental princiis induced in the conductor. of the

Principles

The ple of all direct- and alternating-current generators. direction of the electromotive force depends upon the direction of

motion of the conductor and the direction of the lines of

force in the magnetic field. directly proportional to the

The magnetude of the e.m.f. is number of lines of force cut per

uniform magnetic field depends is moving and the angle upon In it is moving through, with reference to the lines of force. of rate and line 344 let direction the the ab Fig. represent motion of the conductor, when it is moving at right angles to the lines of force (across the magnetic field). When moving at a different angle, such as that shown at &'&', the conductor

unit of time

;

this for a given

the rate at which the conductor

will not cut so it

many

moves at the same

lines of force in a given time, rate.

Here

it

although has moved only a distance

at right angles to the lines of force equal to the projection of the line a'~b on the horizontal as indicated by the line eft', f

which represents the rate at which the lines of force are cut for this direction of motion through the magnetic field. When the conductor is moving parallel with the lines of force as

ELECTRICAL MACHINERY

312

not cutting any lines of force; congenerating no e.m.f. From this it is obvious

represented by a"&", it is

it is

sequently that the e.m.f. generated by a conductor moving at a constant rate in a uniform magnetic field depends upon the angle at which the conductor is moving with reference to the lines of force allel

varying from zero when the conductor is moving parwith the lines of force to a maximum when it is moving

at right angles to them.

Alternating Electromotive Force.

When

a conductor

is

revolved about an axis at a constant rate in a uniform magnetic field, its rate of motion is the same with refe/ence to the

FlG. 345

FlG. 344

Shows how voltage

FIGS. 344 and 345.

lines

is proportional to of force cut in unit time.

number of

neutral axis of the magnetic field as that of the body moving across the diameter C in Fig. 340. In Fig. 345 the con-

A

ductor

c starts

from the position

A

and revolves around

in the

dotted circle at a constant rate in the uniform magnetic field between the north and south poles. When it is in the neutral

AB

moving parallel with the lines of force; consemotion with reference to the horizontal axis of the magnetic field is zero; but as it moves around in its path, its motion across the horizontal axis increases until it is in the axis

quently,

it is

its

vertical axis

XY,

horizontal axis,

at

and

which point

it is

moving

parallel to the

at right angles to the lines of force

and

consequently generating the maximum e.m.f. Beyond this point the e.m.f. decreases until it again becomes zero when the neutral axis is reached at B. On the other half of the revoluis

tion the e.m.f. generated in the conductor will pass through

PRINCIPLES OF ALTERNATING CURRENTS the same series of values as on the

first half,

posite direction, for the conductor is netic field in the opposite direction.

313

but in the op-

moving across the mag-

From

the foregoing

it

motion of the conductor across the horizontal axis in Fig. 345 is the same as the motion of the body across the diameter of the semicircle in Fig. 340, or in other

will be seen that the

TABLE

Ang.

II

NATURAL SINES AND COSINES

ELECTRICAL MACHINERY

314

words, the conductor has harmonic motion across the magnetic Since the e.m.f. is directly proportional to the motion

field.

of the conductor across the magnetic field at

any instant the

electromotive forces generated at the different instances are sometimes spoken of as harmonic electromotive forces. It

the

was shown in Fig. 341

maximum

.that,

where the radius equaled body moving along the

rate, the rate of the

diameter at any instant is equal to the vertical distance between the two bodies for a corresponding period and proportional to the sine of the angle between the two bodies. If the radius about which the conductor is revolving in Fig. 345 is let equal the maximum rate at which the lines of force are cut,

and therefore the maximum

e.m.f.,

the vertical distance

between the conductors and the horizontal axis at any instant will equal the rate at which the lines of force are cut, or value of the e.m.f. generated for a corresponding position.

FIG. 346.

Shows how

sine curve

is

developed.

In Fig. 346 if several positions are taken as shown, to indicate the location of the conductor at different points in its revolution a curve may be constructed as was done in In this case, however, the vertical distance between the curve and the base line will represent the value of the e.m.f. generated by the conductor for a corresponding instant.

Fig. 343.

by a conductor revolving about an axis at a constant speed in a uniform magnetic field is the ideal pressure curve from an alternator and represents quite class closely the e.m.f. and current waves generated by a large

The

sine curve generated

of alternating-current generators.

\

PRINCIPLES OF ALTERNATING CURRENTS

315

The line AO, Fig. 347, is used to Electrical Degrees. denote time, the period of time being expressed in degrees, minutes and seconds. It may at first be difficult to understand how time can be expressed in degrees, minutes and seconds instead of hours, minutes and seconds; but this is easily explained when it is considered that our 24-hr, day is the

time required for the earth to complete one revolution (360 deg.) on its axis. When it has revolved through 15 deg., then 15

/36o,

or

V

of a

24

day

(1

hr.),

If the time

has passed.

required by the earth to make a revolution or part of a revolution can be expressed in degrees, it follows that the time required by any device to perform its cycle of operation may be expressed in degrees, minutes and seconds.

50 60'90

ltd

1

fei

"___ / Cycle

J-

-

FIG. 348

FIG. 347

FIG. 347. FIG. 348.

Eepresents an alternating current or voltage cycle. Eepresents the magnetic field of four-pole machine.

One should be

careful not to confuse this with geographical

or space degrees, although they are the same for a two-pole machine. In a four-pole machine the conductor would make

only one-half revolution, that is, revolve through 180 space degrees to pass by a north and one of its adjacent south poles, as shown in Fig. 348, where the conductor starts from the position

X on the

axis

AB

and revolves

to the position Y.

has generated a cycle, as in Fig. 346

but in this In doing so it case the conductor has passed through only 180 space degrees, although it has generated a complete cycle of 360 electrical or time degrees.

On

;

the other half of the revolution another

generated, the curve in Fig. 349 indicating the complete series of values the e.m.f. passes through on a four-pole cycle

is

ELECTRICAL MACHINERY

316

machine, which shows that the conductor in making one revolution (360 space degrees) generated 720 electrical or

time degrees. In a six-pole machine the conductor would pass through 120 space degrees, or one-third of a revolution, in generating 360 electrical or time degrees; in an eight-pole machine, 90 deg.

etc.

I

2 Alternations .

H 4 Alternations

>|

2 Cycles

FIG. 349.

->J

Kepresents two cycles of alternating current or voltage.

an alternating current in a circuit is explained and 352. Beginning at A on the time line, the voltage starts from zero and increases in value Fig. 350,

The flow

of

in Figs. 350, 351

until

it

reaches a

maximum

at 90 deg.

Assume

this gives the

alternator a polarity as indicated then the Fig. 351 increasing e.m.f. will set up an increasing current in the circuit of a direction as indicated by the arrowheads, and its in

;

be represented by the dotted curve, Fig. 350. Be90-deg. point the voltage decreases, becoming zero at the 180-deg. point; likewise, the current. At this point the

value

may

yond the

conductor passes under poles of opposite polarity and the This will change the polarity of the e.m.f. is reversed. machine as in Fig. 352, and again the e.m.f. and the current pass through a series of values from zero to

from maximum

maximum and From this

to zero, in the opposite direction.

seen that the current and voltage are continually surging back and forth in the circuit; a wave in one direction being called an alternation, and a wave in one direction and then in

it is

the opposite direction, a cycle, or period, as shown in Fig. 347. An alternating current may be likened to the ebb and flow of

the tide, the ebb being one alternation, the flow tide another alternation, the two combined forming a cycle.

PRINCIPLES OF ALTERNATING CURRENTS Another analogy

is

shown

317

Suppose the

in Fig. 353.

pis-

ton to start from rest at the position shown and gradually increase in motion and come to rest at the other end of the cylinder.

pipe from

This would cause the fluid to flow through the If the piston reverses and returns to its to A.

B

would flow in the opposite direction. completed a cycle, and if the piston is kept forth and the water will be kept flowing back and back moving

original position, the fluid

Thus the

fluid has

forth in the system. This is the way a current of electricity flows back and forth in an alternating-current circuit.

FIG. 351

FTQ. 350

^^,

>.

g| ^Jp

Direction of Current for

-

Cjjrve ,

FIG. 352

FIGS. 350 to 353.

A Number of

FIG. 353

Kepresent the flow of alternating current.

Conductors Connected in Series Generating

Voltage. What has been shown of the action of one conductor moving around in a magnetic field is also true of a number of conductors connected in series, as in Figs. 354 to 357.

Here, for purposes of illustration, 'a ring armature is several coils in series connected to two collector

shown with rings, r and

r', the arrowheads indicating the direction of the induced in the armature conductors for the direction of rotation and the polarity shown. When in the position indicated in Fig. 354, the voltage between the collector rings will

e.m.f.

be zero, as the e.m.f. generated in the section ab opposes that in section ad, likewise, the e.m.f. in section &c opposes that in section cd. Consequently the e.m.f. between points Z> and c will be zero,

and these are the points

rings are connected.

to

which the

collector

ELECTRICAL MACHINERY

318

When

the armature has moved around to the position in Fig. 355, the electromotive forces generated in the conductors are in the same direction on either side of the

shown

points where the rings connect to the winding; that

FIG. 354

the

Fia. 355

Fia. 357

FIG. 356

FIGS. 354 to 357.

is,

Diagrams of a single-phase

alternator.

voltages generated in each conductor on either side of points a and c are in the same direction, and they combine to produce

a

maximum

e.m.f. of

a direction as shown.

When

the arma-

ture is in the position of Fig. 356, the e.m.f. will again be zero for the reason described in connection with Fig. 354.

Hence, the position

e.m.f. at the collector rings starts

shown in Fig.

position in Fig.

beyond which

it

354, increases to a

from zero for the

maximum

at the

355, then decreases to zero for Fig. 356; increases until the armature is in the position

PRINCIPLES OF ALTERNATING CURRENTS

319

of Fig. 357, where the voltage between the collector rings is maximum, but in the opposite direction to that in

again at a

When the armature completes the Fig. 355, as indicated. other quarter of its revolution, it will again be in the position shown in Fig. 354, and the voltage will be zero consequently ;

the e.m.f. has passed through the same series of values on the last half of the revolution as it did on the first half but in the opposite direction, just as that generated by the conductor in Fig. 346.

Most modern alternating-current machines above 37.5 kw. are constructed with the armature winding stationary

and the

field

structure revolving

and are known as revolving-

field alternators.

At

may seem

that, since the voltage of the constant and that of the alternatingcurrent machine constantly changing, there must be some first

thought

it

direct-current machine

is

radical difference in the construction of the

two machines.

However, the only essential difference is that the directcurrent machine requires a commutator, for changing the alternating e.m.f. generated in the armature coils into a direct e.m.f. at the brushes. The relation between a direct voltage and an alternating one will be understood by referring to Figs. 360 to 361, which show the same combination of coils and polepieces as in Figs. 354 to 357, except in the latter case the coils connect to a segmental commutator instead of two continuous rings. The brushes B and B', Fig. 360, rest on segments a and e, which connect to the points of maximum e.m.f., giving the external circuit the polarity indicated. As the armature continues to revolve, segments a and e move out from under the

brushes and segments b and / move in, as in Fig. 359. This still leaves the brushes in contact with segments on the same axis as in the first case, which is also true in Fig. 360. From this

it

will be seen that the brushes are always in contact with

a position in the armature coils of maximum voltage and constant polarity, which will deliver a continuous current in the external circuit, or, as

it is

usually called, a direct current.

320

ELECTRICAL MACHINERY

It will be observed from the foregoing that the voltage of a direct-current machine having the same winding as a singlephase alternating-current machine will generate a constant

maximum e.m.f. of the altenating-current indicated in Fig. 361, the curve representing the alternating voltage, whereas the vertical distance between

voltage equal to the

machine.

This

is

FIG. 360

FIG. 359

FIG. 358

FIGS. 358 to 360.

Diagrams of a direct-current generator.

the time line and the line

BC

equals the direct-current voltage.

apparent that a given armature winding will generate a smaller effective e.m.f. when connected to two slip rings than when connected to a commutator in other words, the alternating e.m.f. will be less than the continuous e.m.f., Hence,

it is

;

the ratio being 0.707 to ...

1.

D.C.Voits

B

g

'O

C

\

30'60 90 WI50l (

Represents a direct-voltage and an alternating voltage.

FIG. 361.

Relation Between Cycles, Poles and Speed. It was shown in the foregoing that each time a conductor or group of conductors passes a pole an alternation is generated; therefore the number of alternations per revolution equals the number of poles. Since one cycle equals two alternations,

it

follows that the

Cycles per revolution

where p

is

the

number

=

alternations per revolution

2 of poles.

If

S

number

22 of poles

p

represents the speed of the armature

PRINCIPLES OF ALTERNATING CURRENTS

321 o

in revolutions per minute, then the revolutions per second will be

The j

.

cycles per second, or frequency, / equals the cycles per revolution

multiplied

by the revolutions per second

(

)

,

or

This is the formula for finding the frequency of an alternator and be resolved into two others one for finding the number of poles

may

where the frequency and the speed are known, which

and another poles

for finding the revolutions per

and the frequency are known, that

is,

is

p=

120 Xf -

;

S

minute where fhe number of

S=

-

.

P Since the alternations per revolution equal the

number

of poles,

it

follows that the alternations per minute will equal the number of poles times the revolutions per minute, or pS. Substituting in formula (1),. alternations per minute

~'

120

whence alternations per minute = 120/.

The

following examples illustrate the use of the preceding formulas: Find the frequency and alternations per minute of a 20-pole alternator running at 150 r.p.m. 1.

The

frequency, / =

=

= 25

-^-

cycles per second.

The

alterna -

minute = pS= 20 XI 50 = 3, 000. Find the number of poles in a synchronous motor running 1,200 r.p.m. on a 60-cycle circuit.

tions per 2.

120/

120X60

S

1,200

=6

poles.

3. Find the speed of an 8-pole induction alternation per minute circuit.

alternations per minute

-5T

at

motor connected to a 7,200

7.200

-s5-

and

S=

120/ =

120X60

-^

^

=900

r.p.m.

The last formula gives the theoretical, or synchronous, speed of an induction motor; the actual speed is less, the

amount depending upon the chapter on induction motors.

slip, as will

be explained in the

CHAPTER XXII

MEASUREMENT AND ADDITION OF ALTERNATING VOLTAGES AND CURRENT Instruments for Alternating-current Circuits. In order measure alternating currents and voltages, ammeters and voltmeters are used as in the case of direct currents and voltages, but the construction of the instruments is different. In

to

Fig. 108, Chapter

VII

current instrument.

is

In

it

illustrated the usual type of direct-

the pointer

P

attached to the

is

coil

A, which is suspended between the poles / and I of the permaWhen current passes through the coil A, it is nent magnet caused to turn because it is in the magnetic field between the poles II. Such an arrangement would not serve to measure an

M

.

alternating voltage or current, because the current passing would be constantly reversing its direction, through coil consequently the coil would tend to swing back and forward in

A

unison with the current.

As

a matter of fact there

would be

because the changes in current are so rapid that the coil cannot follow them before it begins to move in one direction, the current through it has reversed and

no motion of the

coil at all,

;

move it in the opposite direction. The objections to the direct-current type of instrument are overcome in one form of alternating-current instrument by the use of two coils instead of a single coil and a magnet. A voltmeter of this type is illustrated in Fig. 362. The coil &

tries to

similar to the one in the direct-current instrument, but in place of the magnet the stationary coils // are used. These are

is

connected together at g and are in series with the resistance R and the coil & across the line, which is connected to the terminals TT.

magnetic

The

field of

coils // are in effect

an electromagnet, the

which reverses for each alternation of the 322

ALTERNATING VOLTAGES AND CURRENTS The coil 6 is situated in this current through them. and the current through it reverses in unison with the Consequently, the force acting to rotate the coil

FIG. 362.

is

323 field, field.

always in

Dynamometer type of instrument.

one direction and will cause

it to move through a greater or depending upon the magnitude of the current. Another type of alternating-current instrument is the one

less distance

illustrated in Fig. 363,

FIG.

soft-iron

vane

363.

b is

known

as the inclined-coil type.

A

Inclined-coil type of instrument.

mounted on a

to the base of the instrument.

shaft,

Vane

which b is

is

perpendicular

mounted on the

shaft at about a 45-deg. angle, and the coil a inclines at a 45-deg. angle with the base. When current. flows through the

ELECTRICAL MACHINERY

324 coil a,

tries to

a reversing magnetic field is set up by it. The vane 6 turn into a position parallel to the field and in so doing

causes the spindle and attached pointer to move. of motion is controlled by a spiral spring.

Voltmeter and Ammeter Readings.

The amount

There can be

little

question as to just what a voltmeter or ammeter should indicate when connected to a direct-current circuit, for the voltage is

approximately constant and the current varies only as the In an alternating-current circuit,

resistance of the circuit.

however, this is not so clearly defined, for here the e.m.f. and the current are not only constantly changing in value, but also in direction. At first thought probably the most logical

answer would be that the instruments will read the average value of the curve. This average value could be found by taking a large number of instantaneous values of the curve,

adding them together and dividing the sum by the number of values taken. Or, it can be shown mathematically that the average ordinate of a sine curve equals 0.636 times the

uaximum

value.

FIG.

if

Thus,

maximum

364.

Shows how values of a

the curve in Fig. 364 represents a voltage whose is 100 volts, the average of the instantaneous

value

values for each half -cycle would be 86.6

-f-

sine curve varies.

+ 50 + 86.6 + 100 +

50 divided by the number of values added together,

namely,

6.

Hence we have, average value of voltage

= 373.2 6

= 62.2

volts.

Owing

been taken, this value is

63.6 volts.

that

to the fact that only a is

However,

not strictly correct. this average value

few values have The exact value is

not the value

would be recorded by an alternating-current voltmeter,

ALTERNATING VOLTAGES AND CURRENTS and

as a matter of fact

it is

325

of no importance in alternating-

current practice and will therefore receive no further attention.

The average value might be indicated on the voltmeter or ammeter were it not for the fact that the effective value of an electric current in a circuit varies as the

times the resistance.

square of the current This effective value of an electric cur-

rent is its heating value, and an alternating current is said to be equivalent to a direct current when it produces the same amount of heat. Since this heating effect at any instant does

not vary as the current but as the square of the current, it follows that to get the heating effect of the current in an alternating-current circuit at any instant, it will be necessary to

square the instantaneous value. Then the average heating equal the average of the square of all the instan-

effect will

taneous values, and the square root of this will give a value equivalent to the direct-current amperes necessary to produce the same amount of heat. Thus, to find the effective value of

an alternating current take the square root of the average of the squares of all the instantaneous values. The effective value is often called the square root of mean squares and for a sine curve

equivalent to 0.707 times the maximum value. stated, the ammeter will read the effective

is

As has been

and not the average value of the current.

The question may voltmeter why the reading on an alternatingcurrent circuit will follow the same law. This will be readily arise

as to

understood when

it is remembered, as explained in Chapter VII, that as far as the indicating device is concerned, it is alike in both instruments; the difference is in the way they are connected in the circuit. The foregoing may be summar-

ized as follows:

The average of an alternating current or voltage sine curve equals the The average value is of very little importance, value XO. 636. as it is not what the instruments read and is therefore seldom used. The effective value of an alternating current or voltage sine curve is the maximum

maximum

valueXO.707, or the the

maximum

voltage

ammeter reading 0.707

=

the voltmeter reading r\~7(Y7

'

an^

ELECTRICAL MACHINERY

326 The

following examples will illustrate the use of these formulas: In a given alternating-current circuit the voltmeter reading is 230, the ammeter reading 75. Find the maximum value of the e.m.f. wave and the maximum value of the current wave. 1.

Maximum Maximum

e.m.f.

current

= =

voltmeter reading

ammeter reading

230

=-

=

75

^-^

= 325 = 106

VO&8.

amperes.

2. Find the ammeter reading and the voltmeter reading of an alternating-current circuit, where the maximum current is 60 amperes and the

maximum

voltage

is

9,350.

Ammeter reading = maximum Voltmeter reading

= maximum

XO. 707 = 60X0.707 =4 -4 amperes. X0.707 =9,350X0.707 = 6,610 volts.

current voltage

Addition of Voltages and Currents.

With

direct current

the resultant of two or more electromotive forces in series

sum

the

of the individual electromotive forces.

When

is

two

alternating-current generators of the same frequency are connected in series, the resultant voltage across the two outside terminals will depend upon the phase relation

sine-wave

between the two machines and will not necessarily be the sum or the difference of the two e.m.f. 's. Consider two alternators A and B, of 115 and 75 volts respectively, connected in series, and let it be assumed that the e.m.f. 's of the two machines pass through their corresponding parts of the cycle at the This condition is instant, as indicated by Fig. 365.

same

usually referred to as the voltages or currents being in step or in phase. In this particular case the voltage across the two machines will be the sum of the two individual e.m.f. 's, or

The dotted curve represents the sum of the instantaneous voltage values and is obtained by adding the instantaneous values on the curve of each individual machine that = ac -f ~bc and od', the maximum of the dotted curve, is, cd 190

volts.

;

equals oaf, the of curve B.

maximum

of curve A, plus ob

r ,

the

maximum

If the voltages of the two machines are in direct opposias indicated in Fig. 366, the resultant effective e.m.f. will be the difference, or 115 40 volts. The dotted 75 tion,

=

curve

is

the resultant e.m.f.

wave obtained by taking the

ALTERNATING VOLTAGES AND CURRENTS

327

d'

~

710*

Vtf 170

300

550 360*.

,

FlG. 370 FIG. 375

FIGS. 365 to

375.

Illustrating the combining of alternating voltages and currents by vector diagrams.

ELECTRICAL MACHINERY

328

difference of the instantaneous values on the curve of each

individual machine

thus od'

;

= oa'

ob'

When

.

two

alter-

nators are connected with the voltage of one directly opposing that of the other, as in Fig. 366, they are said to differ in phase by 180 deg.

In practice, however, the electromotive forces of the individual machines are not always in step, or in direct opposi-

hence the resultant voltage or current will not necesFor example, sarily be the arithmetical sum or difference. consider two alternators in series with their e.m.f. 's in the relation indicated in Fig. 367. In this case the e.m.f. 's do not tion;

pass through their corresponding periods at the same instant, that of machine reversing before that of machine B. From

A

positions

e

to / they are in the

same direction; therefore the

Instantaneous values between these two points will be the sum of the instantaneous e.m.f. 's of the individual curves.

A

At / the e.m.f. of machine reverses and is in opposition to that of machine B until point g is reached, where the e.m.f.

B becomes zero consequently from f to g the resultant instantaneous values will be the difference between of machine

;

those of the individual machines.

From g

to

In

the e.m.f.

's

are

During direction, therefore are added together. the last period of the cycle from h to i they are again in opposition, and the resultant will be their difference. in the

It circle

same

was shown in Chapter XXI that if the radius of the through which a conductor is revolving in a uniform

magnetic field is taken to represent the maximum electromotive force, E, the vertical distance between the conductor

and the neutral motive force,

e,

axis represents the instantaneous

in

any

position.

This

is

electro-

again indicated in

Fig. 368, or in simpler form by Fig. 369, which is called a "vector diagram." Here represents the angle through which the conductor or group of conductors has revolved with reference to the neutral axis. It was also shown in Chapter XXI that the sine of an angle of a right triangle equals the opposite side divided by the hypotenuse. Then in Fig. 369

sine

$= E~;

hence

e

=E

sine



which means that the value

ALTERNATING VOLTAGES AND CURRENTS of a sine

wave

equals the

of electromotive force or current at

maximum

329

any instant

value of the curve times the sine of the

angle through which the conductor has passed.

To

mum

illustrate this, in Fig.

e.m.f. is

315

370

it is

assumed that the maxi-

volts.

Find the instantaneous value of e at 50 deg. Hence sine of 50 deg. is 0.766 (see Table II, Chapter XXI). This is shown on the curve, Fig. 370, and 6 = 315X0.766=241.3 volts. (1)

The

the vector diagram, Fig. 371.

Find the instantaneous value of e at 155 deg., Fig. 370. 155 deg.) sine of 155 deg. is the same as the sine of (180 deg. = sine 25 deg. =0.423. Therefore, e = 315X0.423 = 133.2 volts, as indi(2)

The

cated in Fig. 370.

The same method applies to a current that is,i = I sine 4> where rent and / the maximum current. ;

sine i is

wave of alternating

the instantaneous cur-

By projecting a' horizontally the instantaneous value maj be shown on the vertical axis X'Y', as indicated by oa in Figs. 371 and 372. Henceforth the instantaneous value will be represented on the vertical axis. Fig. 373 is a vector diagram showing the relation between the two e.m.f. 's at the 50-deg. point previously represented by

The line oa', Fig. 373, represents the maximum pressure of machine A, and ob' the maximum of machine B. Since the voltages are in step, they will coincide, and the Fig. 365.

vertical

machines

distances oa

A

and ob

B

and

will represent the e.m.f. 's of If oa' and ob'

respectively at this instant. in the vector diagram are made to equal oa' curve, Fig. 365, then

oa,

and ob' on the and ob in the vector diagram will

equal ac and be respectively on the curves, Fig. 365. Such a diagram is known as a "polar vector diagram/' It is defective in that it does not

maximum

or instantaneous values.

which overcomes vector diagram.

"

this defect

Here the

show the resultant of the Fig. 374

is

a diagram

known as a "topographic two maximums are combined in

and

is

their proper phase relation and where the two e.m.f. 's are in step they will fall in the same straight line and in the same ;

ELECTRICAL MACHINERY

330

direction, the resultant being the arithmetical sum of the two. Applying this diagram to the conditions considered in Fig. 365, we have oa' , Fig. 374, corresponding to od' Fig. 365 '

;

likewise for the instantaneous values ob

equal be and ac, Fig. 365, respectively. ob may be found as follows: .:^

Maximum Therefore, the

e .m ./.

= effective

and ba, Fig. 374, will The values of oa and

e.m.f. .

maximum e.m.f. of machines A and B,

= 162.6

volts

and0.707

0.707

= 106

Fig. 365, will equal

volts, respectively.

In Fig. 373 oa = oa'Xsine 50 deg. As oa' = 162.6, and sine 50 deg. = = 162.6X0.766 = 124.7 volts. Also, o6 = o&'Xsine 50 deg. = 0.766, then oa 106X0.766 = 81.2 volts. In Fig. 374 b'a' = oa Fig. 373, hence oa', f

,

= 106+162.6 = 268.6 volts; and oa = oa'Xsine Fig. 374, equals ob'+b'a' 50 deg. =268.6X0.766 = 205.8 volts. This checks up with the sum of 06, Fig. 373, which is 124.7+81.1=205.8 volts. Fig. 375 is a polar vector diagram of the e.m.f. 's of the generators Since the voltages are in direct referred to in connection with Fig. 366.

oa and

which represents the maximum e.m.f. of B, will be drawn which represents the maximum e.m.f. of A. In this diagram, as in Fig. 373, there is no line representing the resultant maximum or instantaneous value. Of course they may be found by sub106 = 56.6 tracting ob' from oa' for the resultant maximum, or 162.6 ob oa for the resultant instantaneous and from value; that is, volts, 124.7-81.1 =43.6 volts. The relation represented in Fig. 367, and again in Fig. 376, may be shown in the same way. Suppose it is desired to find the resultants at the instant when A has passed through 50 deg., as indicated on the curve; since B is 30 deg. behind A, 50 deg. on A will be 20 deg. on B. This condition is indicated in the vector diagram, Fig. 377, where oa' is drawn to scale to represent the maximum of A, making an angle of 50 deg. with the horizontal axis XY, and ob' is drawn to scale to represent the maximum of B, 30 deg. behind oa'. The distances oa and 06 on the vertical axis will represent the instantaneuos values of A and B respectively, and oa' may be combined with ob' to show the resultant maximum and instantaneous values, as in Fig. 378. Here oa' and ob' have been drawn as in Fig. 377, and then a'd" is drawn parallel to ob', and b'd' parallel to oa', forming the parallelogram oa'd'b' with a'd'=ob' and b'd' = oa'. The distances oa and ad will equal the voltages of A and B respectively at this instant, and od their resultant, which is equivalent opposition

ob',

180 deg. from

oa',

ALTERNATING VOLTAGES AND CURRENTS to cd, Fig. 376. and voltage of

A

The diagonal od

B

The maximum Figs.

f

will

equal the resultant

331

maximum

corresponding to od' on the dotted curve, Fig. 376. and are the same as in e.m.f.'s of machines

The

365 and 373.

resultant

A

B

maximum

od',

may

be calculated

from the formula, (od'}

2

= (oa'} 2 H- (a'd

Angle U = 180 deg. -angle 30 deg., or -0.866.

6

f

2 )

- 2oa X a'd' X cos f

U.

= 180 deg. -30 = 150 deg., and cos

150 deg.

= -cos

FIG. 378

FIG. 377

FIGS. 376 to 378.

Addition of two alternating voltages on currents

when they are out of phase. Substituting the numerical values,

(
volts,

approximately.

Again, '

cos x)

ELECTRICAL MACHINERY

332 and cos

x

2Xod'Xoa' (260)

+ (162.6)

2

2

-(106)

2 =

0.980.

2X260X162.6 Referring to Table II (Chapter XXI) it will be found that the cosine =0.982 and that of 12 deg. =0.978. As 0.980 lies between these

of 11 deg.

values, the angle x will be 11 deg. 30 min.,

and angle # = 30

deg.

11 deg.

30 min. =18 deg. 30 min.

From

the foregoing it is seen that the resultant of oa' and ob f falls This is 11 deg. 30 min. behind oa' and 18 deg. 30 min. ahead of ob' the position of the dotted curve relative to the curves A and B, as shown '.

in Fig. 376.

The

resultant instantaneous value

od = od'Xsin (y+z)', that

is,

e=E

of od

sin (y+z).

is

found by the formula

Substituting the numerical

6=260X0.6225 = 161.85 volts, approximately. The method used in finding the resultant maximum and instantaneous

values of

may be applied in finding the resultant effective value. Also, inasmuch as the effective value of a sine curve equals the maximum times 0.707, the effective voltage across the two machines may be obtained by multiplying the resultant maximum, as found in Figs. 376 and 378, by 0.707. The resultant maximum in Figs. 376 and 378 is 260 volts. The voltmeter reading #X0.707 = 260X0.707 = 183.8 volts. The combining of two sine curves produces a third curve, which is itself a sine values

curve.

CHAPTER XXIII TWO-PHASE AND THREE-PHASE CIRCUITS. Two-phase

Systems.

The

alternating-current

circuits

So dealt with in the preceding chapters were single-phase. for was used current as lighting puronly alternating long single-phase system gave complete satisfaction. advent into the power field, however, the difficulty arose of making the single-phase motor self-starting without

poses,

With

the

its

employing some auxiliary starting device at the expense of ruggedness and simplicity of construction. Ferarris in 1888 discovered that a rotating magnetic field could be produced by two or more alternating currents displaced in phase from one another. This lead to the develop-

ment

of the polyphase systems Instead of revolving one conductor about an axis in a magnetic field, as in Figs. 345 and 346, Chapter XXI, let two .

conductors be arranged 90. deg. apart, as in Fig. 379. It is obvious in this arrangement that when conductor A is on the vertical axis generating a maximum e.m.f., conductor B will be on the horizontal, or neutral axis and its e.m.f. will be As the conductors move in the circular path as indizero.

B

A

will decrease and that in cated, the e.m.f. generated in increase until they are in the position shown in Fig. 380, will be a will be zero, while that in where the voltage in

B

A

maximum.

From

seen that the e.m.f. 's generated in A and B will pass through the same series of values, but when and one is at a maximum the other will be at zero. Curves this

it is

A

which B, Fig. 381, are said to be 90 electrical degrees apart. To produce this combination of e.m.f. 's the conductors will be 90 space degrees apart only on a two-pole machine; on a four-pole machine

show the

relation between the

333

two

e.m.f.

's,

ELECTRICAL MACHINERY

334

they will be 45 space degrees apart

on a six-pole machine 30

;

deg. apart and on an eight-pole machine 22.5 Fig. 382 shows the same arrangement of ;

deg. apart.

and polein but this case four as in 354 to pieces 357, Chapter XXI, Figs. collector rings are used instead of two. If the four rings are concoils

nected to the winding at equidistant points around the armature, in this case 90 deg., when the armature is in the position indicated, a

maximum e.m.f

.

will be generated

between collector

rings a and a' and zero between & and &'. When the armature, has revolved through 90 deg. to the position indicated in Fig. 383, the e.m.f. between rings a

a' will

FIG. 380

FIG. 279

FIGS.

and

379 to 381.

Illustrate

81 F.G. 381 in

voltage

zero, while that

between

b

have decreased to

an elementary way how a two-phase is

and

generated.

&'

will

have increased to a max-

imum. Three-wire Two-phase Systems. From the foregoing it is 's may be taken from the same winding, which are 90 electrical degrees apart. This is the same as was generated by the two conductors, Figs. 379 and 380; therefore curves A and #, Fig. 381, show the relation between the e.m.f. 's taken from collector rings in Figs. 382 and 383. In practice, instead of using a ring armature with one evident that two e.m.f.

winding as in Figs. 382 and 383, two windings are used similar to the arrangement shown in Fig. 384, which represents a four-pole two-phase alternator. In this winding there are 16

coils

phase.

per

circuit, or as it is usually

One phase

of the winding

is

termed, 16

coils

per

shown shaded and the

TWO-PHASE AND THREE-PHASE CIRCUITS

335

The windings are so distributed that when A is under the polepieces and will there-

other unshaded.

the winding of phase fore be generating a

maximum e.m.f., that of phase B will be between the polepieces, consequently the e.m.f. in it will be zero; hence the e.m.f .'s generated by these two 'windings will be the same as those generated by the ring armatures, Figs. 382 and 383.

FIG. 383

FIG. 382

FIGS. 382 to

383.

Diagrams

of two-phase alternators.

For simplicity the conventional diagram

of Fig. 385

may

be used to represent the two phases, indicating that the two windings are displaced 90 electrical or time degrees on the armature. more convenient way is shown in Fig. 386,

A

especially where the windings are grouped. So far only the four-wire two-phase system has been considered. When a winding is used for each phase, as in Fig. 384, they may be connected in series and a common terminal

brought out, as in Fig. 387, which

is

known

as a three- wire

This would be impossible with only one any two of the rings were together it would

two-phase system. winding, for

if

short-circuit the winding. Since the e.m.f .'s of the deg., the voltage

two phases are out of step by 90 between the two outside terminals, Fig. 387,

sum but the resultant, as exFig. 388 is a simple vector diashowing the relation between the two e.m.f. 's of a two-

will not be their arithmetical

plained in Chapter XXII.

gram phase

circuit.

336

ELECTRICAL MACHINERY

TWO-PHASE AND THREE-PHASE CIRCUITS

By adding the two e.m.f.'s vectorially, the OACB, Fig. 389, is formed, the diagonal OC

337

parallelogram

being the re-

sultant of the two voltages and representing to scale the In pressure between the two outside terminals in Fig. 387. this case since the two e.m.f.'s are equal, the parallelogram will be a square, and the diagonal of a square equals one side multiplied by \/2- If it is assumed that the effective e.m.f. of each phase is 135 volts, the pressure between the two outside 191 volts approximately. 1.414 terminals will be 135

=

X

mm Phase ase A

4

Phase B

Wire 2 Phase

2 Phast FIG. 385

FIGS. 385 and 386.

F.G. 386

Four-wire two-phase diagrams.

The same relation holds with the current that is /' /-y/2, where /' is the current in the middle wire and / that in either outside leg, where the current of each phase is the same, assume the current per phase, Fig. 387, to be 35 amperes, the current in the middle wire is I' I \/2= 35 X 1.414 49.5 ;

=

=

amperes.

The foregoing statements may be verified by combining the two curves in Fig. 381, as indicated by the dotted curve, which obtained by adding algebraically the instantaneous values on the curve of each phase. The maximum of this resultant

is

curve

is

1.414 times that of

A

or

B; furthermore,

the dotted

and displaced from A and B, being 45 deg. behind 45 deg. ahead of B. Since the current in the center wire of a balanced threewire two-phase circuit is 1.414 times that carried by either of curve

is

A

the outside wires, the copper in the center conductor will have to be 1.414 that of one of the outside conductors. From this

evident that the saving in copper in the three-wire system over that of a four- wire is not as great as might be expected

it is

;

ELECTRICAL MACHINERY

338 that

is,

the copper in the former

is

85 per cent of that in the

latter.

The expressions E' = E,V72 and across, or the current in,

/'

= / \/2 hold true only where the e.m.f

each phase

is

the same.

In Fig. 390

if

.

the

E

A be represented by EI, that of phase B by 2 and the f then E' = VE^+E^, which voltage between the two outside legs by the of two the where phases are not equal. To can be used voltages find the current in the center leg where the current in the two phases voltage of phase

E

,

A

Phase B

Phase A

J

5 UJ

I 2

Phase FIG. 388

FIG. 387

Phase B

Phase A

'-=27/

FIG. 390

FIG. 389

FIGS.

387 to 390.

Three-wire two-phase diagrams.

the foregoing may be changed to /' = V/^+Za 2 where the current in the middle wire and h and h that in phases A and B

differ in value, /' is

,

respectively.

the three-wire two-phase circuit, Fig. 390, the 235 and that of phase B 175 volts, and the current in phase A 56 amperes and in phase B 43 amperes, let it be required to find (1) the e.m.f. between the two outside terminals and (2) the current

Assuming that

voltage of phase

A

in

is

in the center leg of the circuit.

= 271 /'

=

= V56 2 +43 = 70.6 2

volts.

amperes.

(1) (2)

TWO-PHASE AND THREE-PHASE CIRCUITS Figs. 39i and 392 are vector diagrams of the voltage ively of Fig. 390.

respect-

A

A

FIG. 392

FIG. 391

FIGS.

and current

339

391 and 392.

Addition of two-phaso voltage and currents by vector diagrams.

In practice the e.m.f. 's are seldom very much out of balance on a two-phase system. Where a lighting system is supplied from a two-phase circuit, the current can be almost any value in the two phases, depending on how the load is distributed, although efforts are usually

made

to

keep

it

as nearly

balanced as possible.

Three-phase Systems. The alternating-current system is used more at present than any other is the three-phase. If three conductors are spaced 120 deg. apart in a two-pole

that

magnetic

as shown in Fig. 393, and revolve about an axis by the dotted circle, each conductor will generate

field,

as indicated

a sine wave of electromotive force just as did the single con^ ductor in Figs. 345 and 346, Chapter XXI. Since the conductors differ in location by 120 deg. the e.m.f. generated in will differ in phase by 120 electrical' or time degrees.

them

Rotating the conductors in the direction of the curved arrow will induce in them, at the instant shown, an e.m.f. in the direction indicated

by the dot and

cross in the center of the

The direction of the representing the conductors. e.m.f. in is up through the plane of the paper, while that induced in B and C is away from the reader. Furthermore, circles

A

conductor

A

at this instant

is

on the vertical axis of the mag-

therefore generating a maximum pressure, while conductors B and C are located in the position between netic field

and

the vertical

is

and horizontal

axis

and are therefore generating

ELECTRICAL MACHINERY

340

an

somewhere between zero and maximum.

e.m.f. of a value

Just what this value

is

can probably best be explained by a

vector diagram.

In Fig. 394 on the vertical axis TY' lay

OA

off

represent-

ing to scale the maximum value of the e.m.f. generated in conductor A. Next draw OB 120 deg. in a clockwise direction

from

OA

making

OB

represent to scale the

maximum

e.m.f.

generated in conductor B. Since B travels through the same path and at the same rate as A, the maximum e.m.f. in B will

OB

equal that in A, therefore

will equal

Illustrate

an elementary way how a three-phase

in

voltage

is

generated.

deg. in a clockwise direction from e.m.f. generated in C.

maximum forth,

OC

120

PIG. 394

FIG. 393

FIGS. 393 and 394.

Draw OC

OA.

will also equal

OA.

OC

OB

to represent to scale the

For the reason already set The positions of OA, OB and of the conductors A, B and C

correspond to the positions respectively in the magnetic field, as represented in Fig. 393. The projections of OA, OB and OC on the vertical axis will represent to scale the value of the e.m.f. generated in the

conductors at this instant, as explained in Chapter XXTT. Since OA coincides with the vertical axis YY its projection f

,

on

this axis will be equal to its total value,

that the e.m.f. generated in

OB

A

which indicates

at this instant is at a

maximum.

YY', distance Ob is obBy projecting value of the instantaneous to scale the this tained; represents horizontally to

TWO-PHASE AND THREE-PHASE CIRCUITS

341

The projection of OC also gives the for the e.m.f. generated in C. This is what would be expected when the positions of and C in the magnetic

e.m.f.

generated in B.

value

Ob

B

are considered, as both are in similar positions but under opposite sides of the polepiece hence at this instant the voltfield

;

age generated in each will be equal.

In Chapter value e

e of

= Max. E

to

OB

deg.

=

XXII

it

was explained that the instantaneous by the formula

the e.m.f. can always be found sine <. In Fig. 394 the value of

:

with reference and OC equals 120 deg. 90 deg. 30 deg. Sine 30 Max. E X 0.5. That is, Ob, which 0.5, hence e <

=

=

)B

^ FIG. 395

FIGS. 395 and 396.

Same

FIG. 396

as Figs. 393

and 394, except conductors are

revolved thirty degrees.

represents to scale the value of the e.m.f. generated in

B

or

C

at this instant, is equal to one-half that generated in A, which is at a maximum value. The voltage generated in and C at

B

same direction, therefore the combined equal and opposite to that in A. For the

this instant is in the e.m.f. of

B

and C

is

direction of motion shown, the e.m.f. in that of C is increasing.

B

is

decreasing while

When the conductors have revolved 30 deg. from the posi' tion of Fig. 393, they will be located in the magnetic field, as indicated in Fig. 395. This brings on the horizontal axis,

B

A

therefore the pressure in it has decreased to zero, and and C occupy similar positions in the magnetic field, but under

ELECTRICAL MACHINERY

342

opposite polepieces consequently the e.m.f generated in them This condition is at this instant will be equal and opposite. shown in the vector diagram, Fig. 396. Since OB coincides .

;

with the horizontal zero,

and

axis, its projection

on the vertical will be

with the value of the

this corresponds

e.m.f. gen-

horizontal. projections Oa and Oc on the vertical axis represent the values of the e.m.f. generated in 60 deg. and C at this instant respectively. Since angle

erated in B.

The

A



=

for either OA or OC the e.m.f. generated in A will equal that generated in 0, but will be opposite in direction. Since = In 60 deg., and sine 60 deg. Max. E 0.866. 0.866, e

=

=

X

other words, the voltage generated in A and C at this instant is 0.866 times the maximum value. Again, it is evident that

sum

of the e.m.f. generated in the conductors polepiece is equal and opposite to that generated

the

under one under the

other polepiece.

a b c

/

FIG. 397.

What

Three-phase voltage or current curves.

has been shown for the two positions of the three

namely, the sum of the e.m.f. 's generated in the conductors under one polepiece is equal and opposite to the sum of the e.m.f. 's generated in the conductors under the conductors

This is illustrated in is true for any position. which shows the curves generated by the three conFig. 397, ductors, Figs. 393 and 395. Since the conductors are 120 deg. apart, the curves will also be displaced from each other by 120 time degrees. Referring to the instant marked a on the other pole

curves, line

it is

seen that curve

when B and C

A

is

at a

maximum

above the time

are at half value below the line, which

TWO-PHASE AND THREE-PHASE CIRCUITS

343

indicates they are opposite in polarity to A, which is just the condition represented in Fig. 393. At the instant marked &

on the curves, which

is

30 deg. from

a,

the value of

B

has

decreased to zero, while that of A has decreased and that of G is increased until they are the same value but in the opposite direction which is the same condition as that represented in Fig. 395. At any instant the sum of the instantaneous values ;

above the line

is

equal to the

sum

of the values below the time

line.

Three-phase Machine with Ring Armature. What has been explained for three conductors revolving about an axis in a magnetic field is also true for three groups of conductors located 120 deg. apart, as represented in the ring armature, Fig. 398. Here the same arrangement of coils and polepieces is used as that for the one- and two-phase machines, but in this instance the

winding

is

connected at three equidistant

Considering the position of the will be seen that the conductors between

points to three slip-rings.

armature shown,

it

AI and A 2 are in a position to generate a maximum e.m.f. Between B^ and B 2 the voltage in a 30 deg. section of the winding above the axis XX' is in opposition to that below this axis, therefore this will leave

only a 60-deg. section of the

winding effective to the right of B 2 to produce e.m.f. between B l and B 2 this is just one-half of that between A t and A 2 at this instant. For the same reason the pressure is only onehalf between C l and C 2 that it is between A t and A 2 This is just what existed in the three conductors, Fig. 393. If the armature is turned through 30 deg. as in Fig. 399, the e.m.f. between B^ and B 2 will decrease to zero, as the voltage generated in one-half of the winding between these two points ;

.

is opposite to that generated in the other half, as indicated by the arrowheads. It is evident also that the e.m.f. between x

A

and

A

2 is

equal to that

between

C and C 2 l

,

but in the opposite

This corresponds to the condition in Fig. 395, also as illustrated on the three-phase sine curves at the point

direction.

marked 6, Fig. 397. The three different groups of conductors on the armature,

ELECTRICAL MACHINERY

344

Figs. 398 and 399, have generated in them the same series of values of e.m.f., therefore the voltage between any two of the brushes will be the same. Hence we have a three-wire system

having three equal voltages displaced in phase by 120 deg. This

just the condition in a three-wire three-phase circuit. In a commercial machine, instead of using a ring windings, is

as in Figs. 398

and

399, three windings are used, one for each

FIG. 398

FIGS. 398 and 399.

FIG. 399

Three-phase alternators.

^hase, distributed in groups according to the number of poles. Jne type of such a winding is shown in Figs. 400 and 401.

These windings have 48 coils grouped for a four-pole machine, which gives four coils per pole per phase; each individual group is lettered A, B, and C respectively. Delta and Star Connections. The scheme of grouping the three phases in Figs. 398 and 400 is called a delta ( A ) or mesh connection. When combined in this manner, they form an equilateral triangle, as indicated in Fig. 402. Frequently the windings are shown diagrammatically connected in the form of a triangle, as in Fig. 403 or as in Fig. 404. At first thought it may appear that the three windings are connected so as to cause a short-circuit.

the case

when

it is

It is obvious that this is

remembered that the

e.m.f.

's

not

in a three-

phase circuit are in such a relation that their algebraic sum at any instant is equal to zero, therefore the potential between points A and C 2 , Fig. 404, is zero. In the delta-connected l

winding

it

is

evident that the voltage between terminals

is

TWO-PHASE AND THREE-PHASE CIRCUITS

.

345

346

ELECTRICAL MACHINERY

TWO-PHASE AND THREE-PHASE CIRCUITS

347

that per phase. The current per terminal cannot be the current per phase, as each leg of the circuit connects to two different phases; therefore the current leaving or entering at

any one

some In a balanced where the current in each phase is

of the junctions between the phases will be

resultant of the current in the two phases.

three-phase system (that

is,

Phase B

itE^S

ELECTRICAL MACHINERY

348'

Connecting

A

2

to

B B 2

as in Fig. 405, changes the relation

by 180 deg. from that in Fig. 404; hence, if OA, OB and OC, Fig. 406, represent the phase relation of the three e.m.f. 's of Fig. 404, OB' will represent the relation of the e.m.f. in phase B to that of phase when conof the e.m.f. in phase

A

nected as in Fig. 405. Here it is seen that the e.m.f. in A and B differ by only 60 deg. instead of 120 deg. combining the ;

two

vectorially, the resultant

E

sents to scale the value of

Connecting by 180 deg.

C2

to

B

2

OA"

obtained, which reprebetween and B l Fig. 405. is

A

,

changes the voltage relation of phase

C

In Fig. 406, if OC represents the position of the e.m.f. in phase C with reference to that in phase B, Fig. 404, OC' will represent the phase relation of the e.m.f. in phase C to that in phase B, Fig. 405, which indicates that the e.m.f. in

C differs in phase by only 60 deg. from that in B. Completing the parallelogram, the resultant OB" is obtained, which represents to scale the value of between the terminals B l and Cj.

E

C2

is

A

also connected to

2

instead of

A

lt

as

was done in Fig.

This changes the relation of the e.m.f. in phase A to that in phase C by 180 deg., and is indicated by OA, Fig. 406, and the resultant OC" gives the value of E between the terminals 404.

A lf This combination gives three equal resultant 120 deg. apart the value of these resultants in terms of E per phase is indicated on the diagram. Such a combination as represented in Fig. 405 is called a 7 (gamma) or star C

l

and

e.m.f.

's

;

connection.

The value of the nection

e.m.f.

between terminals with a star con-

equal to the voltage of one phase times the square root of three. If # represents the e.m.f. per phase, then E\/% represents the voltage between terminals. It is evident that is

the current per terminal in the star connection is the same as that per phase as indicated in Fig. 405. The star connection is sometimes represented by the conventional Fig. 407. Fig.

401 shows

how

star connection.

to

group the three windings of Fig. 400 in a practical treatise on wind-

For a complete

ings for alternating-current machinery the reader to

is

referred

Mr. A. M. Dudley's admirable book, "Connecting Indue-

TWO-PHASE AND THREE-PHASE CIRCUITS tion Motors,"

from which Figs.

384, 400

349

and 401 are taken.

Mr. Dudley's book is published by the McGraw-Hill Book Co. Alternating-current generators may be wound for any number of phases, but in practice they are never constructed for more than three. Alternating-current sytsems of more than one phase are referred to as polyphase systems, such as, a two- or a three-phase system.

<E ->:>

Fia. 408

FIGS. 408 to 410.

FIG. 409

Fia.

410

Show

voltage relations on an Edison three-wire, a three-wire two-phase and a three-wire three-phase circuit.

In general there are three types of three-wire circuits: (1) The Edison three-wire system, where the voltage between the two outside legs of the circuit is twice that between either outside and center leg, as in this system may be either direct or alternating current. (2) three-wire two-phase system, in which the potential between the two outside terminals is that between the outside and center terminal times the square root of two, as indicated in Fig. 409; such a system can be

Fig. 408;

A

alternating current only. delta- or star-connected;

(3)

The

three-wire three-phase system, either

combination the voltage between any two terminals is the same as shown in Fig. 410, and can be alternating current only. Obviously either system may be easily identified by the For a determination of the power in voltages between its terminals. in this

alternating-current circuits, see Chapter

XXV.

CHAPTER XXIV OPERATING ALTERNATORS IN PARALLEL. Condition for Parallel Operations.

When

attempting to

operate alternators in parallel, it is necessary to observe the same precautions as in the case of direct-current generators, namely, that their voltages are of the same value and in the

same

direction.

The

latter of these requirements

is

more

diffi-

cult of fulfillment in the case of alternators, since the direc-

tion of their voltages is continually changing. To make clear the relations that exist, it will be convenient to refer back to

the parallel operation of direct-current generators. Suppose that two such generators are connected together as illustrated in Fig. 411. When the polarities are as shown,

each generator will deliver half of the power if they both generate the same voltage. Suppose now that the generators have been connected with opposite polarities, as shown in Fig.

A

very large current would then instantaneously flow the machines by way of the leads and J5, but no through current would be delivered to the load through the leads C

412.

A

and D.

As a matter

short-circuit.

We

of fact, the condition existing would be a might illustrate the conditions of Figs. 411

and 412 by hydraulic analogies, as in Figs. 413 and 414. In Fig. 413 we have two centrifugal pumps that are rotated in the same direction and deliver the same pressure. The flow of water will be from a and through P t to the pipes p, and thence by way of P 2 back to b 1 and Z> 2 Suppose now, that pump 2 .

were to be connected in reverse, as indicated in Fig. 414. It would in consequence pump water from a 2 to & 2 so that all the water pumped from & t to a x through pump 1 would pass >

through pump 2 and return to & x to go through the two pumps again and again. That is, water would merely be circulated 350

OPERATING ALTERNATORS IN PARALLEL

351

around between the two pumps without causing any flow of water through the pipes p. Let us assume that the rotary pumps are replaced by

o

plunger pumps, as illustrated in Figs. 415 and 416.

If both

pistons are traveling from left to right as indicated arrows in Fig. 415, water will be forced from A^ and

A

P

x,

through p to

P

2

and back into B^ and

B

2.

On

by the 2

into

the other

ELECTRICAL MACHINERY

352

moving from right to left when piston from left to right, as in Fig. 416, the space A 2 moving would increase as fast as A l decreased, and similarly B would if

hand,

piston 2 were

1 is

l

B of A

increase as fast as

water

is

forced out

forced from

B

Consequently, whatever would flow into A 2 and all the water

decreased.

2

l

The same conditions would exist on the return stroke, except that the flow would be in the reverse direction that is, from A 2 to A and from B^ to B 2 The result is that the water is transferred from one pump to the other, but none flows through p. The foregoing examples, Figs. 415 and 416, may be likened to the behavior of two alternators in parallel. Thus, in Fig. 417 we have two alternators connected in multiple to a load 2

would flow into B^.

;

.

FIG. 417.

L.

If

cycle

Simplified diagram of two alternators connected in parallel.

a v and a 2 are both positive, and at the same stage of a that is, if the alternators are in phase we would have

the condition of Fig. 5 and current would flow from a t and a 2 to L and back to b^ and 6 2 during one-half of a cycle and in

the reverse direction during the other half. From the foregoing it is apparent that to prevent shortcircuits between alternators when connected in parallel, they

must be connected together when in phase, and that their frequencies must be the same at that time. We therefore have three things to observe when about to connect an alternator in as parallel with others, namely, that its voltage is the same that of the mains, that it is in phase with the main voltage, running at the same frequency as the rest of the the syssystems. Once an alternator has been connected to all under the of that system, tem, its frequency will be slow to tended mover ordinary conditions. Thus, if its prime down, the alternator would act as a motor and take power

and that

it is

OPERATING ALTERNATORS IN PARALLEL from the other machines

to cause the

prime mover

to

353

maintain

On

the other hand, if the prime mover tried to speed up, it would tend to increase the frequency of its alternator, and this it could do only by increasing the frequency of its

speed.

all

the other machines connected in parallel with that alter-

nator.

Adjusting Load on Two or More Alternators. The fact that all the machines on a system run in unison at all times, is 1

the cause for a feature in the parallel operation of alternators that differs widely from the similar operation of direct-current generators, namely, the manner in which the load on the

system

is

divided

among

the machines.

When

it is

desired

that a direct-current generator shall take a greater proportion of the load on the system, its field flux is increased by cutting resistance out of its field rheostat, thus causing an increase in

generated voltage and a corresponding increase in the load The increase in load results in a slight slowing down

current.

of the prime mover, sufficient to cause the governor to admit

the additional steam or water required by the added load. An attempt to increase the load of an alternator in the

The foregoing manner would, however, result in failure. speed of the machine being fixed by the frequency of the system to which it is tied, an increase of field current can only increase the generated voltage, but cannot make the machine slow down, and hence cannot make the governor admit more

steam. Consequently, the alternator cannot assume more of the load, and the field-current adjustment -does not change the load distribution. It is found, however, that such an increase in generated voltage causes a current to circulate

between the

generator of higher voltage and those of a lower voltage, thereby causing a greater current to flow through them than that required

by the

load,

and consequently causing an un-

necessary loss in the machines.

It is therefore a part of cor-

rect parallel operation to see to it that all the like excitation.

machines have

The prime movers of alternators should be equipped with governors that can be adjusted while the machines are in

ELECTRICAL MACHINERY

354

Then, when the load is to be increased, the goveris adjusted slightly, causing it to admit more steam or water, as the case may be. This will tend to cause the machine to increase in speed. However, a very slight increase in speed will serve to make the alternator take a greater share of the operation.

nor

load than before, and so reduce the load on the others. Putting an Alternator Into Service. As pointed out in

the foregoing that in order to operate two or more alternators in parallel, it is necessary that they be of the same voltage, of the same frequency, and that they be in phase. The first of these requisites is readily determinable by means of a voltTo determine the other two, several devices may be meter. employed. Of these the simplest is that in which two ordinary

incandescent lamps are used.

FIG. 418.

Conn

In Fig. 418 are two single-phase

synchronizing

,-

lamps alternators, 6r

L./O

and G 2 with 6^ connected ,

alternators;

dark.

to the busbars B.

Suppose that it becomes necessary to put G 2 into service also. Having started G 2 and brought it up to normal speed, its voltage is adjusted to be the same as that of the mains. At switch $ 2 lamps L : and L 2 have been connected in the manner

shown and these lamps should nicker. If the two alternators are almost at the same speed the flickering of the lamps can readily be followed by the eye; they will slowly light up to full brilliancy and then as slowly dim down to complete dark-

up again. This will continue with perfect regularity so long as the speeds of the generators remain ness, only to light

OPERATING ALTERNATORS IN PARALLEL

355

exactly as they are. Any change in speed, however slight, causes a change in the intervals of brightness and darkness of

the lamps.

Having found that the lamps flicker, the speed of G 2 is changed slightly and the effect on the lamps noted. If the flicker becomes more rapid it is an indication that the speed has been changed in the wrong direction. If it becomes less rapid, it is known that the speed is more closely approaching 6r x Finally, when there is no flicker it is proof that the two speeds are exactly the same. If the lamps are burning brightly when the flickering ceases, it is a sign that G 2 is 180 deg. out of phase with G i and that a short-circuit would occur

that of

.

$ 2 were closed, even though the generators are running at the same speed. If the lamps are burning dimly, it is an indication that the machines are out of phase by some angle less than 180 deg., under which condition it is still not permissible

if

$2

to close

.

When G 2

is

in phase with

G

lt

the lamps remain

they die down to darkness and then do not again light up, it is known that the two machines have been brought into phase and it is safe to close switch 8 Z9 thereby This process is frequently connecting G 2 in parallel with 6r dark.

Thus,

if

.

referred to as synchronizing the two machines and are in phase they are said to be in synchronism. It is

when they

not usual to attempt to get the absolutely ideal condiby the lamps not lighting up at all. Ordinarily,

tion indicated

the speed adjustment is made until the lamps are flickering very slowly; that is, until it takes several seconds for the

lamps to light up, dim down, and light up again. We know then that the speeds of the two machines differ only very The lamps are watched as they slowly grow dimmer slightly. and dimmer and finally become entirely dark. During this period and before they again begin to light up, the switch is

82

closed.

Sometimes the lamps are connected as in Fig. 419 then the instant of synchronism is that at which the lamps are at full brilliancy just the opposite from Fig. 418. This method ;

has the advantage that

if

one of the lamps should happen to

ELECTRICAL MACHINERY

356

burn

become manifest by the lamps refusing of speed by the operator, and he will not be led to believe that the machines are in synchronism, as might be the case if the dark method were out, the fact will

to light

up under any manipulation

' '

* *

On the other hand, the precise instant of synchronism can be so much more readily judged in the case of the dark

used.

method, and the likelihood of an operator being deceived by the accidental burning out of a lamp is so slight that the "dark" method is in greater use than the "light" one.

FIG. 419.

Connections for synchronizing two single-phase alternators;

lamps bright.

The reason that the lamps guide

is

Fig. 418, erator GI

very simple.

When

method provide a

in either

terminal

c

of generator

G

2,

positive at the same instant as terminal a of genand the voltage of the two machines are of the same

is

value, then there can be

no flow of current from one machine of one machine completely

to the other, since the voltage

neutralizes the voltage of the other so far as a circuit through the two machines is concerned. Therefore lamps L v and L 2 will not light up, indicating that the two machines have the same polarity. However, if terminal c is positive when terminal a is negative, then current can flow from c around to a through the armature of G^ to b back to d and through armature G 2 to c, thus completing the circuit. It is seen from this that the same conditions exist as would if two direct-current machines of opposite polarity were connected through lamps Lj and L 2 double potential of one machine would be applied

OPERATING ALTERNATORS IN PARALLEL

357

across the lamps, and if they are for the voltage of one machine they will come up to full brilliancy, indicating, when connected as in Fig. 418, that the alternators have opposite polarity when the lamps are bright. In Fig. 419 if terminal

G2

same instant as a of 6r 15 and the voltage of both machines is the same, then the two machines are in synchronism. However, in this case current can flow from terminal c of G 2 through L 2 around to b through G l to a and back through L^ to d and through G 2 to c. Consequently the lamps will burn bright when the machines are in sync of

is

positive at the

chronism.

Other Methods of Synchronizing.* The earliest method used to determine the condition of synchronism was by connecting lamps in series between the machines, but is unsatisfactory because they can do nothing more than to indicate a difference in voltage above that value required to glow their Voltmeters have been substituted for lamps, but filaments.

while they indicate any voltage difference, they do not show the relative speed of the machines or the phase angle between them. Only in the synchronoscope synchronism indicator are

all

these desirable features combined.

Whatever

synchronoscope is used, the generator the synchronizing instrument's pointer is traveling slowly and is close to the zero. There are a number of makes of synchronoscopes, or synchronizers. The

switch

is

style of

closed

when

Lincoln synchronoscope, Fig. 420,

is

essentially

a bipolar,

split-phase, synchronous motor operating on the difference in phase between the machines being synchronized. Its two field coils, F and F, are connected in series across one phase of the bus, on the running machine, consequently excited with alter-

nating current instead of direct current. Its two armature coils, A and #, located at an angle of 90 deg. to each other, are

connected in parallel through slip rings and brushes S to the corresponding phase of the incoming machine. The armature

and *

field cores are of

laminated iron.

The remainder of this chapter appeared in Power, Sept. 16, 1919, title "Methods of Synchronizing Alternating-Current Machines,"

under the

by Frank

Gillooly.

:

' I

',

,1

,

I

ill

I I

'

:,

;

,

111

'

i

,

,

i,

;

,i

.

I

1

;.

::

J

.,

i;

;

,

"

!

',

.i

I

i

i

, i

J

the field circuit tion where

As

its

this point

Fb\ causes

Hie

amp a*

lines of Force <M.'iacide

changes the armat \Poirrfer

Bus FIG. 420.

Machine Schematic diagram

o:i

ncoln synchronoscope.

field coils lags the voltage across them by nearly 90 deg., and the current in the armature coil B, due to the inductance in series with it, lags the voltage across it by approximately 90 deg., at exact synchronism B will be in the

current in the

position where its field will be parallel to the field of FF. This position of the armature at synchronism brings its

pointer to the vertical position, as shown in Fig. 420, corresponding to the indication on a clock at the hour of twelve.

The indicator rotates clockwise when the incoming machine "fast" and counter-clockwise when the incoming machine "slow."

is

is

OPERATING ALTERNATORS IN PARALLEL

359

Another type of synchronoscope operating on the same 421. The moving principle is shown diagrammatically in Fig. element consists of two vanes, A and A, on either end of a cylindrical iron core in the center of the fixed coil C and fastened to the shaft carrying the pointer B. The rotating and situated produced by the two stationary coils through a resist90 deg. apart and connected in parallel,

M

field is

N

M

ance

R

and

N

through an inductance

/.

Synchronoscope similar to Fig. 420, but with

FIG. 421.

all

coils

sta-

tionary.

The vanes are excited alternately positively and negatively by the alternating field of coil C and are deflected by the fields has its current in phase with that of of coils and N. Coil C and deflects the vanes so as to parallel the field of with

M

M

M

that of C.

Coil N, with the inductance 7 in series, has

rent lagging that of

C by

its

cur-

approximately 90 deg. and so long which is the point ;

as this phase relationship is maintained

of synchronism

it

exerts

no torque on

A

and A.

At any

ELECTRICAL MACHINERY

360

N

N

will deflect the other angle of phase between and C, vanes; and with a difference in frequency between machine and bus a varying torque will be produced that will rotate the shaft.

In this instrument the rotating field is produced in the winding by the bus potential and the alternating field by the machine potential, an arrangement opposite to split-phase

that of the Lincoln instrument, Fig. 420. Hence, the indicator's direction of rotation is counter-clockwise when the in-

coming machine is "fast" and clockwise when the incoming machine is "slow." Electrodynamometer Type of Synchronoscope. The synchronoscope shown in Fig. 422 operates on the electrodynamometer principle, in which the torque exerted on a movable coil A, suspended between two stationary coils F and F, is opposed by spiral control springs. The stationary coils F and are connected in series through the resistance R F across the bus or running machine. The movable coil A, carrying the pointer, is connected in series with a condenser C and a resist-

F

RA

A

across the machine to be synchronized. 90-deg. phase displacement is thus obtained by making the current in lead that in and by means of the condenser.

ance

,

F

A

F

At exact synchronism

the true 90-deg. phase angle be-

tween the movable and stationary coil circuits no torque is exerted on A, and it is held at zero position in the center of

A difference in frequency by the control springs. between the machines will produce a varying torque, and A If the will oscillate the pointer back and forth on the scale. frequencies of machine and bus are equal but there is a difference in phase, A will take up a position of balance along the

scale

scale corresponding to this

phase angle.

zero, central, position of the scale

Right and

are "fast"

left of

the

and "slow"

respectively with regard to the machine. The pointer merely oscillates back and forth on the horizontal scale instead of

rotating over a circular dial. The dial is illuminated by one lamp connected to the secondary 8 of a transformer having

two primaries

P

and P, connected, one across the incoming

OPERATING ALTERNATORS IN PARALLEL

361

machine and the other across the bus, and is bright at synEach revolution of the synchronoscope pointer represents one cycle difference in frequency between the alternators. The deflection of the pointer from zero at any instant represents the phase angle between the machines. chronism.

is not intended to be kept energized not in use or to be revolved at too great a speed. Where synchronizing lamps are provided, the speed of -the alternator

The synchronoscope

when

should be adjusted closely by them before the synchronoscope

TbMacftine

FIG. 422.

Synchronism indicator that operates on the electrodynamometer principle.

plugged in circuit if necessary, the synchronism indicator be plugged in for a moment in. order to determine whether the incoming machine is fast or slow. Just as the lamps can be checked against the synchronoscope when synchronizing, so should the instrument be checked against the

is

;

may

lamps when phasing out apparatus. Synchronoscope Connections. In Fig. 423 is shown a simple form of Lincoln synchronoscope installation that is frequently met with. The synchronizing bus runs to all genAll receptacles are wired alike, to the same erator panels. phase on each machine. One side of the generator potential transformers

is

grounded, the other sides are brought to the

ELECTRICAL MACHINERY

362

bottom terminals T^ and plugs are used:

One

P

r,

T2 is

of the receptacles. Two kinds of inserted in the receptacle of the

running machine, the other, Ps in the receptacle of the incoming machine. The top connection, B and B 2 of each receptacle goes to conductor B of the synchronizing bus connected ,

,

to the field coils of the synchronoscope.

each receptacle,

M

i

and

M

2,

goes to the

The third point conductor

M

of

of the

Fuses

To

FIG. 423.

Diagram of connections

Genenarfor No.B

for Lincoln synchronoscope.

The bus connecting to the armature of the synchronoscope. fourth terminal of the receptacle is not used with grounded secondary transformers.

shown

at

XL,

of which

The inductance-resistance box

lamp

G

is

the resistance.

Lamps L

is

are

on the base of the synchronoscope and are connected so as to be

dark at synchronism.

CHAPTER XXV KILOWATTS, KILOVOLT-AMPERES AND POWER FACTOR.

How Current Builds up in a Direct-current Circuit. In a direct-current system, if a constant pressure of E volts is applied to a circuit having R ohms resistance, the current /, in amperes, that will flow in the circuit will equal -^

ohms'

1=

p<

R

ohms

(Ohm's Law). is

For example, when a

resistance of 1.5

connected across the terminals of a 105-volt direct-

current generator, the current set

=

Or

up

in the system

is

_=

/= _ R

70 amperes, which is the value of the current after o it has reached a steady flow. fact frequently overlooked is that the current does not instantaneously attain a constant -L

A

flow,

but gradually increases to a normal value.

This

is illus-

trated in Fig. 424, where the current in amperes and the pressure in volts have been plotted on the vertical ordinate to the same scale and the time in fractions of a second along the

Starting at the point marked* zero, which represents the closing of the circuit, the full pressure is instantly applied as shown by the line marked "volts." However,

horizontal.

instead of the current at once reaching a normal value, it gradually increases, in the first tenth of a second reaching a

value of 43 amperes, in two-tenths about 60 amperes, in threetenths 65 amperes, and in one second approximately normal value, or 70 amperes. At this point the current curve becomes parallel with the voltage.

This gradual building up of the current 363

is

similar to start-

ELECTRICAL MACHINERY

364

ing a machine with a heavy flywheel, during the period of acceleration of which energy is being stored up in the moving mass. Just as energy is stored in the moving mass, energy is also stored in

an

electric circuit.

In the

latter,

however, the

storing of energy while the current is increasing is due to the magnetic field set up by the current and is more pronounced in a circuit consisting of a coil of wire wound on an iron core than in a straight conductor or coil in open air. After

the machine has been accelerated, if the source of driving power be removed, it will continue to keep in motion for a

HS

FIG. 425

To D.CLint

r 01

02 03

04 05

0.6

0.7

SECONDS

0.8

09

1.0

LI


L-J 11-?

KILOWATTS, KILOVOLT-AMPERES, POWER FACTOR of the windings.

365

Similarly, a machine with a heavy flywheel

cannot be brought to rest suddenly without wrecking the machine. If an ordinary incandescent lamp be connected in parallel with a coil of wire

the switch

is

wound on an

iron core, such as a

terminals of a generator, and closed to a source of an electric current, as in

lamp connected across the

field

Fig. 425, a current will be set

up in the system as indicated. After the current has reached normal value, if the switch be opened, the lamp will continue to burn for a short period, because of the energy stored in the magnetic field, although disconnected from the circuit. The direction of the current

through the lamp and

coil,

after the switch

is

opened

is

indi-

cated in Fig. 426. From this it is evident that the current does not suddenly come to rest, but has a tendency to keep in

motion just as the machine did with the flywheel.

01

0.2

03

0.4

05 06 07

0.8 0.9

SECONDS FIG. 427.

10

Fig. 427,

I.

Curve showing how current decreases in a

circuit.

what takes place in the circuit described in Fig. 424, when the source of electromotive force is removed; instead of

illustrates

the current coming to rest instantly, shown by the curve.

Lenz's Law.

The foregoing

is

it

gradually decreases as

evidence of what

is

called

first stated by Lenz and known as "Lenz's Law." Briefly stated, this is in part: "Whenever the value of an electric current is changed in a circuit, it

the law of inductance,

ELECTRICAL MACHINERY

366 creates

an electromotive force by virtue of the energy stored

in its magnetic field, which opposes the change/' as explained in Chapter XII.

up

Current Lags Behind Voltage.

What

has been shown in

Figs. 424 to 427 is that the current lags behind the electro-

motive force, reaching its normal or zero value an instant later than the voltage. With direct current this occurs only upon

opening and closing the switch or when the resistance of the circuit is changed, such as when devices are added or removed from the circuit; therefore it would not be expected to seriously affect the flow of the current. However, in an alternating-current circuit where the current is changing back and forth very rapidly, it might be expected that this property (inductance) would have some effect upon the flow of the current. This statement is borne out by referring to Fig. 424. Here it will be seen .that if the circuit is allowed to remain 1

/ 10 of a second, the current would increase only to about 43 amperes before it would decrease again. Fig. 428 shows the effect of inductance in an alternating-current closed for only

The full-line curve represents the impressed voltage whereas the dotted curve indicates the current that passes through its maximum and zero values an instant later than

circuit.

At the point b the electromotive but the current in the circuit is of considerable value and does not decrease to zero until point c is reached. the electromotive force. force

is zero,

Between these two points the electromotive-force curve below the time

line,

while the current curve

is

above.

is

This

indicates that the current

is flowing opposite to the voltage of the generator, which does not seem in accordance with the law governing the flow of current in a circuit.

Consider what takes place in an alternating-current circuit: In Fig. 429 a generator G is connected to a load L. Assume the electromotive force to be building up a polarity as shown. When it has reached a value indicated by a on the curves, Fig. 428, the current is zero beyond this the current increases in value in the same direction as the electromotive ;

force.

This continues until the instant & on the curves

is

KILOWATTS, KILOVOLT-AMPERES, POWER FACTOR

367

reached; here the electromotive force of the generator has zero, but not so with the current. "When the current

become

begins to decrease, it sets force which tends to keep

up

in the circuit an electromotive

flowing just as in the direct-current circuit. At the instant the voltage of the generator becomes zero, the electromotive force induced in the circuit is of a value and direction such as to keep the current flowing. it

Thus the conditions are reversed the generator becomes the load and the load the generator. From points 6 to c on the curve the load sets up a current through the generator, as

FIG. 428

Fia. 429

L

FIG. 430

F.G. 431

FIGS. 428 to 431.

Illustrate a lagging current.

indicated in Fig. 430, just as the coil in Fig. 426 sets up a current through the lamp after the switch had been opened. Also, from points b to c the voltage due to inductance is decreasing while that of the generator is increasing in the

At c the voltage of the generator has opposite direction. increased to a value equal and opposite to that due to the inductance, and the current becomes zero. Beyond this point the current reverses and again flows in the same direction as the generator electromotive force. This is what happens at

each reversal of the current. Figs. 429

and

from the generator but generator.

The ammeter

in the circuit,

430, not only reads the current that

From

this it is

is

flowing

from the load to the obvious that the ammeter reading

also that flowing

ELECTRICAL MACHINERY

368 will be too high

where inductance

is

present in the circuit, and

this will be referred to later.

Pump Analogy

of

Alternating

Current.

A

physical

analogy of the action of an alternating current in a circuit is given in a valveless pump, Fig. 431. If the piston be oscillated back and forth in the cylinder an alternating current of

water will be produced in the system. The pressure applied to the piston not only has to overcome the resistance due to the friction of the water on the sides of the cylinder and pipe, but has also to overcome the inertia of the water, first in getting it started from rest at the beginning of the stroke, and again

it

has to bring

it

to rest at the

end before

it

can start

in the opposite direction. This is similar to what took place in the alternating-current circuit, Figs. 429 and 430. The electromotive force of the generator not only had to overcome

the resistance of the circuit, but also the back pressure induced by the changing value of the current.

In the water analogy it would not be correct to say that the total force applied to the piston is useful in keeping the elements in motion. At the beginning of the stroke work

had to be done in getting the mass in motion, and again at the end of the stroke in bringing the mass to rest; consequently the power supplied to the piston may be divided into two parts a useful component overcoming the friction of the

system and keeping the mass in motion, and another that is used in overcoming the inertia of the elements. This latter component might be considered as not doing any useful work, for at the end of the stroke it has to undo that which it did at the beginning.

In an alternating-current circuit, as the current increases therefore it produces a back pressure in the circuit

in value

;

the total electromotive force produced by the generator is not effective in overcoming the ohmic resistance of the circuit, but

part of it is used in overcoming the inductance. Similarly, on the latter half of the alternation as the electromotive force decreases to zero the current does not come to rest with sets

it

but

up an electromotive force which tends to keep the current

KILOWATTS, KILOVOLT-AMPERES, POWER FACTOR

369

flowing; consequently, the generator will have to apply pressure in the opposite direction to bring the current to rest. Hence it would not be proper to consider that the total pres-

sure

is

useful in setting

made up

up a current

in the circuit, but instead

two components one a useful component, which overcomes the ohmic resistance of the circuit, and another component which overcomes the inductance. Apparent Watts and Useful Watts. In a direct-current it

is

of

system the volts multiplied by the amperes equals the watts transmitted in the circuit. This is also true in an alternatingcurrent circuit where the current is in step with the electromotive force, as shown by the curves, Fig. 432.

Curves of voltage and current in step.

FIG. 432.

such a condition

However,

is

rarely true in practice, the current being out of step with the volts. It has been

usually more shown that in an alternating-current circuit where inductance is present, the voltmeter and ammeter readings give too high a value for the effective volts and amperes, therefore the product of these two readings will give too high a value for the useful watts. For this reason the product of the volts -multiplied by the amperes in an alternating-current circuit is or less

"apparent watts," or volt-amperes. A wattmeter will read the useful watts, or "true watts," transmitted irrespective of the relation between the currents and electro-

called

motive force.

A Wattmeter Reads True Watts. Why a wattmeter will read the useful watts transmitted in the circuit will be evident to Figs. 433 and 434. Assume the current and electromotive force to be building up a polarity as indicated in Fig. 433, the current having a direction through the wattmeter elements as shown by the arrowheads. While the cur-

by referring

ELECTRICAL MACHINERY

370 rent

increasing, part of the energy is stored up in the magand is not available for doing useful work. Inas-

is

netic field

much

as the wattmeter registers the product of the volts and amperes, it will therefore be registering too high a value for the useful watts on the first half of the alternation. As the

current dies to keep

it

down

in the circuit, the effect of inductance

is

flowing.

In the Figs. 428 to 430 it was shown that when the electromotive force of the generator decreased to zero the current did not, but for a period the load supplied current to the gen-

The

erator.

upon the wattmeter

effect of this condition

^WATTMETER

.WATTMETER

FlQ. 434

FlQ. 433

FIGS. 433 and 434.

is

Watt-hour meter connected on single-phase

circuit.

shown in Fig. 434, from which it will be seen that the current through the movable element is reversed from that in Fig. 433, but is still flowing in the same direction in the current coils. Reversing the current in one element only will reverse the torque of the instrument, that is, it will have a tendency to turn backward. The backward torque continues from points b to c on the curves, Fig. 428, which is just enough to offset the excess reading of the meter on the first half of the alternation, beyond point c the current builds up in the same direction as

the electromotive force and the process

Power

in

an Alternating-Current

is

Circuit.

repeated.

The product

of

volt

by amperes (apparent watts) in an alternating-current circuit usually called "volt-amperes," and is equivalent to the actual watts a direct-current circuit. In direct-current work the number of kilo-

multiplied is

in

watts

is

expressed

by the formula, kw. =

.

J.

(kv.-a.) is

The term

kilo volt-amperes

jUvJU

used in alternating-current work and equals

volt-amperes .

-L

is

general practice to rate

all

It

jUUu

alternating-current generators, synchronous

KILOWATTS, KILOVOLT-AMPERES, POWER FACTOR

371

motors, synchronous condensers and transformers in kilovolt-amperes instead of kilowatts. The ratio of the apparent watts to the true watts is

called

the

"power

factor."

By

transposition four other

formulas are obtained, namely: Watts volt-amperes X power factor; volt-amperes = watts -f- power factor; kilowatts = kilovolt-amperes X power factor; and kilovolt-amperes = kilowatts -f-

power factor.

The

following problems will illustrate the application of the foregoing

formulas

:

Find the power factor of a single-phase alternating-current circuit when the voltmeter indicates 450 volts, the ammeter 50 amperes and the wattmeter 20 kw. 1.

Volt-amperes

Kilovolt-amperes

= 450 X 50 = 22,500 :

1,000

kilowatts

Power factor =

=-

kilovolt-amperes (or

What

.

= volt-amperes = 22,500 =22.5. 1,000

2.

(apparent watts)

size of generator

is

20 =0.89.

22.5

89 per cent) required to supply a load of 35 kw. at

85 per cent power factor? Kilovolt-amperes

kilowatts

=

power factor

=

35 =41.2. 0.85

The power factor

is always less than one, except when the in step with the electromotive force. Under such a condition the true watts and apparent watts are equal, which

current

is

is referred to as unity power factor. In the second one of the is effects of low power factor problem manifested, namely the generator has to be of larger capacity to supply the load when the current is out of step with the electromotive

condition

force than when they are in phase; therefore it is desirable to keep the power factor as near unity as possible. Effects of Capacity. So far only inductance has been mentioned as affecting the relation of the current to the electro-

motive force.

There

is,

however, another element present

known

as capacity, or in other words, the property of a con-

denser.

The effect of capacity in an alternating-current cirthe same as inductance only in the reverse order; that

cuit

is

ELECTRICAL MACHINERY

372 is,

inductance causes the current to lag behind the electro-

motive force, while capacity causes the current to flow ahead of the electromotive force.

Capacity has the same effect in a circuit as would be experienced if a chamber C with an elastic diaphragm across it were placed in the pipe line, Fig. 431. Fig 435 shows this

pushed toward one end of the cylinder, the diaphragm will be distorted as indicated by the dotted line, to accommodate the water displaced from the cylinder. Distorting the diaphragm creates in it an effect against the motion of the piston, so that when the piston is reversed, If the piston

arrangement.

is

the energy stored in the diaphragm will assist in reversing

FIG. 435.

it.

Hydraulic analogy of capacity (condenser effect) in an alternating-current circuit.

The energy stored in the diaphragm is opposite to that stored up in the moving mass due to inertia. If the mass did not possess inertia, or the energy stored in the mass were less than that stored in the diaphragm, the piston and water would reverse sooner than the force applied to keep it in motion.

Capacity has a similar effect in an alternating-current circuit, by causing the current to reverse before the electromotive force, just as inductance prevented the current from changing its

direction in the circuit until a very short interval after

the electromotive force

The

had changed

its

relation between the current

direction.

and electromotive force

where capacity only is present is represented in Fig. 436. Here it will be seen that the current passes through its zero and maximum values an instant earlier than the electromotive force and is said to lead the voltage. The power in a circuit

factor of such a circuit

is

In Fig. 435,

if

power factor, while power factor. distorting the diaphragm

called a leading

that of an inductive circuit

is

called a lagging

the pressure due to

KILOWATTS, KILOVOLT-AM PERES, POWER FACTOR

373

equal to the inertia of the mass, the water and piston will move back and forth in time with the force applied. Similarly, in an alternating-current circuit, if the inductance and capacity are made equal, the current will flow in step with the electromotive force, as indicated in Fig. 432, and artificial means are frequently used to obtain this end. is

Power-factor instruments have a scale similar to that in Fig. 437, unity power factor being indicated at the

shown

When

the needle deflects to the right, it indicates a leading power factor; to the left, a lagging power factor. The power factor equals the cosine of the angle of lag or lead center.

between the current and the electromotive the angle by which the current

is

force.

Therefore

out of step with the electro-

ELECTRICAL MACHINERY

374

and capacity in the circuit is called reactance, inductance being referred to as inductive reactance and capacity (or condenser effect) as capacity reactance. The combined effect of resistance, inductance of the circuit.

Power phase

and capacity

is

called the

impedance

In the foregoing singleof Polyphase Circuits. If been is let to have considered. only

E

circuits

represent the volts and / the current per phase in a balanced two-phase circuit, that is, a circuit in which the volts and

amperes of each phase are equal, the Kv.-a. =?

= _LI^WI

*? Jf T ZILI..

and

in a

1,000'

1

balanced three-phase circuit Kv.-a.

'

1,000 (1)

If

the voltmeter reading in a two-phase circuit is 2,300, the kilowatts, find the kilo volt-amperes

ammeter 50 and the wattmeter 180 and power factor.

~

2EI

and the power

The voltmeter

_180_

reads 6,600, the

meter 0.85 on a two-phase

&

=230knovolt-amperes,

factor

to.

(2)

2X2,300X50

2EI ~' :=

1000

circuit;

ammeter

15,

and the power

factor

find the kilowatt load.

= 2X6,600X15 =198kilovolt-amperes. 1000

Kiv.=kv.-a.XP. F:

= 198X0.85 = 166.3

kilowatts;

the wattmeter reading. (3) In a three-phase circuit the wattmeter reads 250 kilowatts, the ammeter reading is 350 and the voltmeter indicates 550 volts, find the

power

factor.

Kv.-a.

=

1.732EI -

= 1.732X550X350 = 333.4 1,000

1,000

to.

250

kilo volt-amperes.

KILOWATTS, KILOVOLT-AMPERES, POWER FACTOR (4)

When

the voltmeter in

ammeter 135 and the power

a three-phase

factor meter 0.90,

circuit

375

reads 2,300 the

what should the wattmeter

head? Kv.-a.

= 1.732EI = 1.732X2,300X135 = 957.8

Kw.=kv.-a.XP.

Where

the load

kilovolt-amperes.

1,000

1,000 F.

is

=957.8X0.90 = 862

kilowatts.

unbalanced on a polyphase circuit take

the average current of the different phases and use this average value the same as the ammeter reading on a balanced

and this will give very close to the correct results in most cases for obtaining the kilowatt-amperes, etc.

circuit,

CHAPTER XXVI POTENTIAL AND CURRENT TRANSFORMERS. The current required as the voltage decreases power that if the voltin Chapter

Potential Transformer Operation.

amount was shown

to transmit a given is

increased.

It

of

XIX

one-half, and the size of the conductor required to transmit a given load need only be one-quarter the size, to carry the current at the higher

age was doubled the current would be

voltage as at the lower voltage with the same percentage losi in the line. This fact is taken advantage of in alternatingcurrent circuits, and by the use of potential transformers the

voltage is stepped up to a high value when power is to be transmitted long distances, and then stepped down to the required voltage at the load. Alternating current is generally

generated at 440, 550, 2,300, 6,600, 11,000 or 13,200 volts. When a higher voltage is required it is usually obtained

through transformers, since it is difficult to insulate rotating machines for a higher voltage than that given in the foreall the parts of a transformer are stationary be they may thoroughly insulated and placed in a tank of oil which also assists in insulating the windings, as well as

going.

Since

acting as a medium to transmit the heat away from the coils. Oil insulated transformers have been designed to operate on circuits

up to 220,000 volts. Types of Potential Transformers.

Transformers

are

designed in three general types, according to the method of In this type the cooling. (1) Oil-insulated and oil-cooled. coils

and core are placed

in a tank of

oil,

and the

circulation

of the oil due to heating is relied upon to transmit the heat from the windings and core to the surface of the tank. (2) Oil-insulated

and water-cooled. 376

"With this type the coils and

POTENTIAL AND CURRENT TRANSFORMERS

377

core are immersed in a tank of oil as in the oil-cooled type, but the tank is made high enough to allow the placing of coils of pipe in the oil at the top of the tank, through which water circulated to cool the oil. (3) Air cooled. In this type the windings and core are mounted in an iron case without a is

bottom; the unit set over a duct and a blower used to blow air

up through the winding, which

the case.

is

discharged at the top of

In this way the heat of the winding and core

is

dissipated.

Principles of Transformer Operation. The principle upon which transformers operate is known as mutual inductance and was explained in Fig. 187, Chapter XII. Here it was shown that if a conductor was in the magnetic field of another conductor carrying a current that was changing in value, the first conductor would have a voltage induced in it opposite to that applied to the conductor carrying the current.

Two

FIG. 438.

coils

on iron

core.

here two coils B is enlarged upon in Fig. 438 and C are wound on an iron core A. Coil C is connected to a source of direct current and at the moment the switch S is closed, a current flows through coil 0, causing A to become a magnet. The magnetic field thus introduced into B generates a voltage in it that can be read on voltmeter V. This voltage

This principle

;

only for the instant occupied to build up the field in A; as the field reaches its final value and stops increasing, the lasts

voltage at

V

opened and

the field in

drops to zero again.

tarily generated in B.

A

When

the switch

8

is

collapses, a voltage is again momenThe reason for this can be explained

ELECTRICAL MACHINERY

378

by reference to Fig. 439, which shows a longitudinal section and core. The magnetic field may be represented broken the lines. It will be seen that they form loops, as by of the coils

at L.

When

the field begins to build up, these loops, beginexpand and in so doing cut across all the

ning at nothing, wires

W

and X.

"We know that when a magnetic flux cuts

across a conductor a voltage is generated in the latter, hence the voltage registered by in Fig. 438. Again, when the field

V

reduced to zero, as by the opening of switch $, the loops of the field in Fig. 439 contract, and consequently cut across the conductors and X, as before, but in the opposite direcis

W

FIG. 439.

Section through coils and core, in Fig. 438.

Hence, the voltage that appears across the terminals of the current through C in Fig. 438 is broken. Since the lines of force, when contracting, cut the conductors in the tion.

B when

opposite direction to that in which they cut them when expanding, it is to be expected that the voltage registered by V when the switch 8 is opened is opposite to that which it regis-

when

is closed, and such is the case. Moreover, found that the direction of the voltage in B is opposite to that in C when the coils are wound in the same direction as they are in Fig. 438. Thus, when a is plus and

ters it

the switch

will be

we find that c is also plus and d minus. Since a voltage is created across coil B, Fig. 438, it is evident that current would flow through a circuit if connected b minus,

POTENTIAL AND CURRENT TRANSFORMERS

379

If the switch 8 were to be continuously closed and it. opened with great rapidity, a pulsating voltage would be established across coil B that would cause a pulsating current It is to flow through whatever circuit was connected to it. therefore apparent that energy from generator G can be transferred to the resistance R, in spite of the fact that no to

electrical

resistance.

we can

connection exists between the generator and the For the continual make and break of switch 8

and

By

substitute an alternating current.

current through

C

will continually be

forth, so that the effect is

of the switch 8.

FIG. 440.

Two

With an

coils

much

made

its

use the

to alternate back

like that obtained

by means

alternating current large values

one wound over the other on an iron core.

can be handled with entirely satisfactory results, and the In practice the core A device is known as a transformer. is extended and bent around to form a complete loop of iron as in Fig. 440

and the

coils

wound on both

legs of the core.

Moreover, it has been found that better operating results are obtained when one coil is wound over the other, instead of side

by

side.

The

result

is

that in actual transformer construction

the coils would be placed somewhat as illustrated in Fig. 440. Core-type and Shell-type Transformers. There are two

types of transformers according to the way the coils and core are arranged; namely, the core-type and the shell-type. In the core-type the coils are placed on the core as in Fig. 441 where in the shell-type the core is built up around the coils in Fig. 442.

For low and medium voltages

either type

may

be

ELECTRICAL MACHINERY

380

used, but for very high voltages most all transformers are of the core type.

The iron

core

is

not

made

of a single piece of solid iron is built up of

bent into shape, as indicated in Fig. 440, but

many

pieces of sheet iron that are

stamped to the shape of the

core, as that illustrated in Fig. 443.

The object of

this is to

reduce to as small a value as possible the currents that are induced in the core, known as eddy currents. Consider a COTS*

Core-

Fia. 441

FIG. 442

FIG. 441.

Core- type transformer.

FIG. 442.

Shell-type transformer.

solid core with a hole

H

in

it,

as in Fig. 444.

prises a complete circuit, as indicated

The core com-

by the arrows.

It is

therefore equivalent to a conductor making a single turn, whose ends are joined. Therefore, if an alternating current flows in coil C, Fig. 444, it will induce a voltage in the core

A, the same as in a turn of any other conductor. Since this single turn is closed upon itself, a current would flow through in Fig. 444 may be imagined to be made it. The hole

H

smaller and smaller, until it closed and made a solid core; however, the eddy currents would still circulate. If the core is

of sheets of iron, as in Fig. 443, it is evident that in circulating around in the core must cross currents eddy

built

the

up

POTENTIAL AND CURRENT TRANSFORMERS

381

from one sheet to another. The resistance introduced in the path of the current by the slight air space that may be between the sheets is

and by the thin coating of

scale

on their

sufficient to restrict the flow of current to

surfaces,

comparatively

small values, whereas the solid core will allow them to assume

FIG. 443

FIG.

444

Shows construction of transformer core. Shows how eddy currents are induced in transformer

FIG. 443. FIG.

444.

core.

values that involve large losses. For simplicity of construction, the individual sheets of iron are not punched out as a

complete loop, as in Fig. 443, but in one type of construction they are made in two pieces and assembled in two parts, as in Fig. 445.

The

coils

FIG. 445.

which are wound on forms

Transformer core in two

can then

sections.

readily be slipped over the two parts of the core, which are then placed together to form a complete circuit of iron. The

appearance of a core and its coils is shown in Fig. 441. Relation of Primary and Secondary Volts. The voltage induced in coil 8 of Fig. 446 when coil P is connected to a source of alternating current, is proportional to the number of turns in the coil. This must be so, since the voltage in 8 is

induced by the

lines of force

which cut across

its

turns as

ELECTRICAL MACHINERY

382

the lines expand outward and collapse with each alternation of the current in P. It is evident that the magnetic flux will

induce the same voltage in each turn of the conductor

it cuts,

and

since the total voltage of the coil is the sum of the voltages induced in its turns, it follows that the greater the number of turns, the greater must be the voltage of the coil,

and

which makes it possible to step voltages. Thus, if coil 8 has 100 turns and the voltage induced in it is 220 volts, we could get 2,200 volts vice versa.

up and

step

It is this fact

down

by substituting a coil of 1,000 turns for 8 and leaving same as it is. Not only is the voltage of 8 proportional

FIG.

turns in

Schematic

446.

it,

but

its

ratio of the turns in

8 has

200,

diagram

the

transformer.

potential

voltage and that applied to P are in the 8 and P. Thus, if P has 12,000 turns and

and 6,600

volts if applied to P, the voltage across

will be 110 volts, because, volts 8 volts 200 6,600 volts 8

=

of

P

to the

PX

12,000

:

volts

X

200

P

:

:

= 110

200

:

volts.

8

12,000, or

This

is

12,000

readily explainable since the voltage applied to P caused a certain current to flow through it. This current creates a flux will which in magnetic every turn of 8 the same produce voltage as exists across each turn of P. The winding to which the source of current

is applied is The transformer. winding in which primary of the a voltage is induced is called the secondary. If the secondary voltage is higher than the primary, the transformer is referred

called the

if the secondary voltage is lower than that of the primary, it is a step-down transformer. Whether it be one or the other, we have the relation, second-

to as a step-up transformer

;

POTENTIAL AND CURRENT TRANSFORMERS dry voltage If

turns.

Ep =

primary voltage

:

we

:

:

383

primary secondary turns by symbols we have, :

substitute the quantities

We

T p where E stands for voltage and T for turns. T, now come to that property of transformers, the posses-

sion of

which makes possible their use for power-transmission

E

s

:

s

:

,

purposes, which is, that the current taken by the primary is automatically controlled by the load that is connected across the secondary.

In Fig. 466,

P

is

the primary

and 8 the

secondary windings of a transformer. How the primary current responds to the secondary load and adjusts itself to the value of the latter may be explained as follows:

The moment current begins to flow through 8, the flux in T will be affected by it. The magnetomotive force set up by the current through 8 will always be in opposition to that created by the current through P. For example, if at a given instant, a is positive and 6 negative, the current the core

through

P

will be in the direction of the arrows,

making the

by the arrowheads on the broken lines representing the lines of force. The voltage in duced in 8 will consequently be in the direction of the arrows direction of the flux that indicated

on winding 8, and the flow of the current will therefore also if switch 8 is closed. The magnetomotive force set up by this current will tend to set up a flux through coil 8, and hence will be opposed to that due to P. The result is that the reactance of P is lowered, since all reactance is due to the flux continually cutting back and forth across the conductors of which the reactance is composed. As soon as the reactance is thus reduced the current through P increases,

be in that direction

until the flux through T is again the same as before. When the current in S is reduced by a change in its load, less flux is developed by it, so that there is less flux to oppose that of This causes the flux through T to increase, thereby inP.

creasing the reactance of P and consequently throttling the current through it to the point where the flux through T

again becomes normal. Grouping of Transformers.

Transformers

may

be con-

384

ELECTRICAL MACHINERY

nected in series or parallel just as batteries are connected in series or parallel. When so connected they are operated on a circuit. For two-phase operation two transformers are used, one on each phase, and can be used separately for four-wire operation as in Fig. 386, or the two adjacent terminals may be connected together, as in Fig. 387, Three transChapter XXIII, for three-wire operation. formers may be grouped in delta by connecting the primaries

single-phase

and secondaries

as in Fig. 404, or

necting the primary

grouped in star by conand secondary windings as in Fig. 405,

Chapter XXIII. Current Transformers. Where large currents are to be measured on alternating-current circuits, or where the current is measured on high-voltage circuits, the instruments are connected into the circuit through a current transformer. This type of transformer is similar in its operation to the potential

transformer but instead of being connected directly across the line as is the voltage transformer, the current transformer is connected in series in one leg of the circuit as in Fig. 447.

POTENTIAL AND CURRENT TRANSFORMERS

385

Standard practice is to design all currrent transformers amperes will be flowing in the secondary circuit with full load on the primary, consequently, most alternatingcurrent instruments and protective devices are designed for so that 5

5 amperes in their current coils. For example, a current transformer designed for 1,000 amperes in its primary would only supply 5 amperes to an ammeter connected to its second1,000 amperes were flowing in the primary. However, most switchboard ammeters are calibrated to indicate the primary current. Such a transformer as referred to in the

ary,

when

5 or 200 The 1. foregoing would have a ratio of 1,000 secondaries of a current transformer should never be left :

:

open-circuited. Before they are disconnected from the instrument or protective device the secondary should be shortIf this is not done, when there is a load on the circuited.

primary the voltage in the secondary will be built up to a very high value, which may break down the transformer's insulation, or the transformer may overheat due to the high iron losses in the core.

CHAPTER XXVII ALTERNATING-CURRENT MOTORS an Induction Motor. Induction motors are two major parts, the stator and the rotor. The made up stator, shown in Fig. 448, is the stationary element and has placed in slots, in the inner periphery of the core, coils which Parts

of

of

are grouped into a winding connected to the power circuit.

FIG. 448.

The

Induction-motor stator.

rotor, Fig. 449, of the squirrel-cage type

is

built

up

of a

outer periphery. In these slots are placed copper bars which are connected to rings at each end of the core, thus forming a closed winding, as shown.

laminated iron core, slotted on

its

In the wound-rotor type the revolving element is wound with an insulated winding and connected to slip rings, Fig. 450, 386

ALTERNATING-CURRENT MOTORS

387

similar to the winding used on a revolving-armature-type alternator. The slip rings are connected to an external resist-

ance used for starting and to control the speed of the motor. The stator winding is the same for both types of machine.

FlG. 449.

Squirrel-cage rotor.

The induction motor

is

differentiated frfim the

current motor in that where the latter

from an outside source

to

both

FIG. 450.

its

Phase-wound

in the induction motor the current

winding only.

direct-

supplied with current armature and field windings, is

is

rotor.

supplied to

its

stator

Then how does the induction motor operate?

the question that naturally arises. This might be answered by saying, On the same fundamental principles as the direct-

is

' l

ELECTRICAL MACHINERY

388

current motor"; namely, the reaction between the magnetic up about the conductors on the revolving element due to the current flowing in them and the flux from the polefield set

pieces, the polepieces in the induction

The fundamental

how

the current

motor being the stator. two machines is in

difference between the

is set

up

in the rotor

winding of the induction

motor, which has no electrical connection whatever with the stator winding. In the stator of an induction motor the magnet field

is

made

to revolve

and

this in

turn reacts upon the

current induced in the rotor bars and carries the latter around with the former.

FIG. 451.

Two-phase voltage or current curves.

Revolving Magnetic Field. The explanation of the revolving magnetic field can in general be most easily given by utilizing a two-phase circuit, the conditions of which may be Instead of represented by the curves A and B, Fig. 451. using a smooth-bore stator with a uniformly distributed winding, a field structure will be

a direct-current machine

employed similar

(see

to that used in

Figs. 452 to 460).

In this

structure there are eight polepieces, four of which are linked and four by winding J5, representtogether by the winding ing the conditions in a two-phase motor. Although there are

A

eight polepieces, lines

it

will be seen in the

following that the

combine in such a way as to form but four discussion it will be assumed that the current

magnetic In this is in step with the voltage, although this is not true in practice it nevertheless has no effect upon the production of the

poles.

;

rotating magnetic

field.

ALTERNATING-CURRENT MOTORS Consider the windings two-phase circuit,

and

A

389

and B, Fig. 452, connected to a A and B, Fig. 451, represent

curves

let

the voltage applied to the windings, consequently the current flowing in them. At the instant a on the curves, Fig. 451, it will be seen that the voltage in phase is at a maximum

A

value, while that in

B

is

at zero, consequently the current in

will be at a maximum value. If the curconsidered flowing in winding in the direction of the arrowheads, poles A^ to 4 will be magnetized of a polarity as

winding A, Fig. 452, rent

A

is

A

FIG. 452

FIG. 452.

FIG. 453.

indicated.

FIG. 453

Schematic diagram of two-phase motor. Same as Fig. 452, but 45 degrees later.

Since the current

field will also

be at a

is

maximum

at a

maximum,

the magnetic

value and will divide at the

N

poles, pass into the rotor core back into the S as in the figure. This division of flux at the illustrated poles,

center of the

pole faces is similar to that in the bar magnet, Fig. 454. Here the flux also divides at the center of the pole and passes around through the air to the S pole.

N

Next consider the condition in the circuit represented by the instant & on the curves, Fig. 451. At this instant the current in has decreased and that in B increased until it is the

A

same value in both.

Applying

this condition to the

motor

windings, the current will be about 0.7 maximum in each and in the same direction as in Fig. 453. This arranges the eight poles into four groups, each containing two like poles. Like

ELECTRICAL MACHINERY

390

N

poles repel, therefore all the flux from any pole will pass into the rotor core and go to the adjacent S pole, as shown in This condition is illustrated with bar magnets in Fig. 453.

A

Here the two similar magnets

Fig. 455.

and

B

are placed

so that their like poles are adjacent to each other, and since like poles repel, the flux from, one pole will be pushed away

N

from that

of the other,

consequently giving a distribution to

the lines of force similar to that

shown

\\

Com-

in the figure.

:N ^;,.

N -v,'

::::

; .::

*V/

/

/

/ \V^!////;/ \lll\\:J^y

FIG. 455

Fiu. 454

FIG. 454. FIG. 455.

Shows magnetic conditions in Fig. 452. Shows magnetic conditions in Fig. 453.

paring Figs. 452 and 454 with Figs. 453 and 455, it will be seen that where the centers of the magnetic fields are at the centers of A polepieces in Figs. 452 and 454, the centers of

A

and B polepieces in the magnetic fields are located between of the magnetic center the Figs. 453 and 455. In other words, field

has been caused to move one-half the width of a pole-

piece.

At the

instant

c,

decreased to zero, and

A

has Fig. 451, the current in phase that in phase B has increased to a maxi-

ALTERNATING-CURRENT MOTORS

391

mum value, consequently the current in the motor's winding A will be at zero value, while that in winding B will be at a maximum The

This condition

value.

is

illustrated in Fig. 456.

A

polepieces has become zero, while that in has reached a maximum value, and the center of

flux in the

polepiece B the magnetic field

is

at the center of the

B

polepieces.

This

bar magnet A, Fig. 455, was removed. Then the flux distribution would be as indicated in Fig. 458, which shows the center of the magnetic field shifted the same as

condition

is

to the left

from between

center of the

B

if

A

and

B

magnets, in Fig. 455, to the

magnet, Fig. 458.

Again, the center of the

B

FIG.

FIG. 456. FIG. 457.

magnetic

field

FIG. 457

456

Same as Fig. 452, but 90 Same as Fig. 452, but 135

degrees later. degrees later.

has been caused to shift one-half the width of a

polepiece.

Returning to the curves A and B, Fig. 451, and considering instant d, it is found that the currents in A and B phases are again of the same value. But curve A is below the line XY, indicating that the current in phase A is flowing in an opposite direction to that in phase B. This condition is illustrated in Fig. 457. Here the current in winding

A

it

will be seen that the direction of

reversed from that shown in Figs. 452 and 453, but the current in B is in the same direction as in This has reversed the polarity of the Figs. 453 and 456. is

A

polepieces, and now the poles of A phase that had in Fig. 452 have become S poles in Fig. 457 and

N

polarity

vice versa.

ELECTRICAL MACHINERY

392

However,

this has again

brought

flux distribution will be as shown.

what would have been the case been taken from the left-hand side

like poles adjacent,

This

is

and the

a condition similar

magnet A, Fig. 455, had B magnet and placed on the right-hand side as in Fig. 459. Again the center of the magnetic field has shifted, this time from the center of the B polepiece to between B and A polepieces.

to

At

instant

e

if

of the

on the curves, Fig. 451, the current in phase

r4s

k

\\M Vn^^-:N \ v \ V \

-!''.'ji/j'

/ ;.|V/ /

i.ii'--

-'y/V-'!V-^

x

/

>-^y

// /

B

ALTERNATING-CURRENT MOTORS

393

from between the B and A poles, Fig. 457, to the center of A poles as shown in Fig. 460. A similar condition would in Fig. 459 if magnet B is removed as in Fig. 461. obtained be center of the magnetic field has shifted from beThere the

the

B

A

magnets to the center of magnet A. Comparing Fig. 454 with Fig. 461, it is seen that the center of the magnetic field has shifted from the center of magnet A,

tween the

and

shown in dotted lines in Fig. 461, to the center of magnet A, shown in full lines, Fig. 461. Similarly in Figs.

Fig. 454,

L..1A

FIG. 461

FIG. 460

FIG. 460.

Same

FlG. 461.

as Fig. 452, but 180 degrees later. Shows magnetic conditions in Fig. 460.

452 to 460, the magnetic

magnets

A

2

and

shifted

A A

4,

from

field

has shifted from the center of

A maximum value N poles, to the center of maximum N poles; likewise the S poles have A and A to A and A

and

l

iwpv i

2

3,

4

x

3.

Relation Between Poles, Frequency and Speed. Referring to the curve, Fig. 451, it will be seen that the current in each phase has passed through the equivalent of one alternation. At a the current in phase A is at a maximum, and that in phase B is at zero value likewise at e phase A is again at a ;

maximum phase

B

value, but in this case in the opposite direction,

is at zero.

and

ELECTRICAL MACHINERY

394

From

the foregoing it is evident that for one alternation the magnetic the motor shifts one pole, then for one cycle (two alternations) the magnetic field will shift two pole spaces. In other words, for each cycle that the current passes through, the magnetic field will shift around

field in

one pair of stator poles. Now, if the motor was connected to a 60-cycle circuit that is, a circuit in which the current makes 60 complete reversals per second, or 120 alternations per second the number of revolutions made by the magnetic field in one second would equal the cycles divided by the pairs of poles. In our problem the induction motor has four poles, or two pairs, therefore the speed per second of the magnetic field 60 =30 revolutions. Looking at this problem for a 60-cycle circuit will be 2 another way, for one cycle the field will shift two pole spaces; therefore, for two cycles, in a four-pole machine, the field will shift four pole spaces or make one revolution; that is, two cycles is equivalent to one revolution. Then for 60 cycles the field will make 60-:- 2 = 30 revolutions, which checks with the value obtained in the foregoing. In one minute the speed will be 60 times that for one second, or in this case,

30X60 = 1,800

P lutions per minute,

the

revolutions per minute.

number

If

S

equals the revo-

and / the frequency,

of pairs of poles,

the cycles per second, then the revolutions per minute S that is, the 1 OA/ speed of the magnetic field is S = Substituting the value 60 for .

the frequency

S

= 120X60

S=

120X25

that

(cycles

=1,800.

= 750

If the

and 4 the number of poles gives

frequency were only 25

revolutions per minute.

a 4-pole motor

if

second)

per

is

From

cycles per second,

this

it

be seen

will

connected to a 60-cycle circuit, its magnetic minute, and on a 25-cycle

revolve at 1,800 revolutions per circuit, 750 revolutions per minute. field will

The formula may be transposed

to read

P=

-

to find the

number

o r> cr

of poles

when the speed and frequency

are known,

and

to deter-

/=

mine the frequency (cycles per second) when the number of poles and speed are known. If the frequency equals 60 cycles per second and the speed 1,800 r.p.m., then

P=

120X60-

=4

poles.

This

is

what

it

should be

1,800 for the

machine

of poles to be 4

in

the figures.

On

the other hand, knowing the number

and the speed to be 1,800

r.p.m., frequency

/=

4X1

800

j-

=60

ALTERNATING-CURRENT MOTORS

395

From this it is evident that there is a fixed relation between the number of poles and the speed of the revolving magnetic field of an induction motor and the frequency of the circuits to which cycles per second.

it is

connected.

How netic

Rotation

field

is

was shown

Produced, to

In the foregoing the mag-

revolve in a clockwise direction.

Starting with this assumption, consider the action of the field on the rotor, to produce rotation. Assume the rotor to be at standstill in Fig. 462, as at the instant of starting, and with the magnetic field revolving in a clockwise direction. The lines of force will be cut in a counter-clockwise direction.

FIG. 462.

Shows

direction of the current in an induction-motor's rotor

conductors.

This condition stationary

and

the same as though the- magnetic field was the rotor was revolved in a counter-clockwise is

This will be made clear by referring to Figs. 463 and 464. In Fig. 463 assume conductor A to be stationary and the magnetic field moved horizontally in a right-hand Then the magnetic field direction, as indicated by arrow B. would be cut by A from right to left. This would be the same as if the magnetic field were held stationary and the conductor moved horizontally to the left, as shown by arrow B, Fig. 464. direction.

When the conductors are cutting the lines of force in a counter-clockwise direction, Fig. 462, those under the poles

N

ELECTRICAL MACHINERY

396

will have a voltage induced in them that will cause a current to flow toward the reader, as indicated by the dots on the end

The conductors under the S poles will have voltage generated in them that will cause current to flow down through the plane of the paper, as indicated by the crosses on the end of the conductors. Considering the action of the conductor.

on the magnetic field produces a turning effort

of the current in the rotor conductors

from the polepieces

will

in a clockwise direction

show that ;

in other

it

words the rotor in an

in-

duction motor turns in the same direction as the magnetic field revolves.

FIG. 463

FIGS. 463 and 464.

FIG. 464

Conductor in magnetic

field.

the instant of starting the rotor is at a standstill and the magnetic field is revolving by the rotor conductor at a

At

speed, as was shown in the foregoing, which equals 120 times the frequency divided by the number of poles, or, for a four-

X

=

60-^-4 1,800 pole machine on a 60-cycle circuit, equals 120 revolutions per minute. This gives the maximum rate of cutting the lines of force, and consequently generates a maxi-

mum voltage in the rotor conductors of a frequency equal to the voltage applied to the stator. the frequency of the rotor current will be the same

Why

as that of the stator current will be understood

when

it

is

ALTERNATING-CURRENT MOTORS remembered that the polarity

of the field coils

397

changed with

each alternation; therefore if the polarity of the field coils changed with each alternation, the voltage generated in the conductors under the polepieces will change with each alternation.

Rotor Must Run Slower than the Magnetic Field. Assume that the magnetic field is rotating at 1,800 r.p.m. and that power is applied to the rotor and caused to also run at 1,800 r.p.m. in the same direction as the magnetic field. Under such a condition the rotor conductors and the magnetic field would be running at the same speed, therefore the former

would not be cutting any lines of force, consequently the voltage in the conductors would be zero and no current would the foregoing it is evident that when the is at a standstill the voltage and frequency are at a maximum, but that as the rotor comes up to speed, the voltage and frequency of the rotor current de-

From

flow in them.

rotor of an induction motor

crease until

when

the rotor

is

running

at the

same speed

as

field, the voltage and frequency become zero. Since no current flows in the rotor conductors when they are

the magnetic

moving at the same speed as the magnetic field of the stator, no torque, turning effort will be produced. In the foregoing is found the explanation why the rotor of an induction motor must run slower than the stator 's magnetic field. The rotor runs just enough slower than the field to set up sufficient current in the rotor bar to develop the torque necessary to drive the load. If there is no load on the motor, then the rotor runs at practically the same speed as the magnetic field, but as the load is increased the speed decreases until at full load

the difference in speed will amount to from about 3 per cent for large, efficient machines to about 10 per cent for small

The difference between the speed of the magnetic and the rotor is called the slip and is usually expressed as a percentage of the speed of the magnetic field, which is the

machines. field

synchronous, or theoretical speed. Where 8 m represents the speed of the stator 's magnetic field and S the speed of the rotor,

per cent

slip

=

Sm

-8

~

X 100.

ELECTRICAL MACHINERY

398 Assume that a

4-pole 60-cycle induction motor runs 1,675 r.p.m. at the per cent of slip. In the foregoing it was found that the theoretical speed of a 4-pole 60-cycle motor is 1,800 r.p.m.; then

full load, find

1

the per cent slip

=

fton

1

f\7f^

XlOO = 7;

that

is,

the rotor

is

running

about 7 per cent slower than the magnetic field of the motor. The speed marked on the name-plate on induction motors is usually the full-load speed, consequently somewhat less than the theoretical speed figured from the number of poles and the frequency.

The current taken from the line by the stator of an inducmotor is not only limited by the ohmic resistance of the

tion

stator winding, which is always comparatively small, but also by the counter-voltage generated in the winding, the latter

being almost equal to the applied volts, there being only sufficient difference to allow the current to flow necessary to carry the load on the motor. The counter-voltage induced in the stator winding is similar to the counter-electromotive force generated in the armature of a direct-current motor. In

the latter case this back pressure is produced by the armature conductors cutting the lines of force from the polepieces,

where in the former

it is

induced in the stator winding by the

alternating magnetic field changing in value about the conductors. At no load the current taken from the line is that

necessary to magnetize the stator and rotor core and supply the losses in the motor. If

we consider

the -action of the current in the rotor bars

alone, as in Fig. 465, field as

shown.

This

it

is

will be

found

to

produce a magnetic

the resultant field from the clockwise

the conductors carrying current away from and the counter-clockwise field set up about the conductors carrying current toward the reader. By comparing Figs. 462 and 465 it is seen that the center of the magnetic field set

up about

the reader

the rotor is located between the poles of the stator. the condition to be desired, since the rotor poles exert a minimum demagnetizing effect on the stator poles, and also field of

This

all

is

the rotor conductors under each pole are carrying current same direction, consequently producing a given torque

in the

with a

minimum

current.

However, owing

to the effect of

ALTERNATING-CURRENT MOTORS

399

inductance in the rotor the current lags behind the voltage generated in the rotor conductors, and instead of the current being distributed as indicated in Fig. 465, the actual conditions approach those of Fig. 466. Here it is seen that, owing

behind the voltage, the rotor's N and S poles come considerably back under the stator's N and S poles respectively. Consequently, the rotor poles have a demagnetizing effect on the stator poles. In fact the rotor current of an induction motor has the same effect upon the

to the rotor current lagging

stator as the secondary

winding of a transformer has on the

primary.

I'^IG.

465.

Magnetic

field set

up by the rotor when the current

is

in step

with voltage.

The Induction Generator. Assume that power is applied and it is caused to run faster than the magnetic field. Under such conditions the rotor conductors to the rotor, Fig. 462,

an opposite direction to that in Fig. 462, consequently the voltage and current set up in the rotor conductors will be opposite to that in Figs. The conditions with the rotor running 462, 465 and 466. will be cutting the stator field in

shown

faster than the magnetic field are

the current set

up

in the rotor

shown in Fig. 467. Now if it is running slower than

when

the magnetic field is a demagnetizing current, the current set up in the rotor when the motor is caused to run faster than

the magnetic field

is

a magnetizing current.

This

is

at once

ELECTRICAL MACHINER

400

Here we find that the N poles of evident from Fig. 467. the rotor are under the S poles of the stator and the S poles under the N poles of the stator. This brings about a condition where the rotor current can supply the

of the rotor

necessary magnetic

field to

generate the counter- voltage in the

However, when the rotor has increased in speed a small percentage above that of the magnetic field the counter-voltage of the stator will have increased above the applied voltage and the motor will become a generator. Then, instead of the machine taking power from the line to drive it, as a motor, it will require mechanical power applied to the rotor to drive the latter faster than the magnetic field, and stator windings.

FIG. 467

FIG. 466

FIG. 466. FIG. 467.

Magnetic field set up by rotor of induction motor. Magnetic field set up by rotor of induction generator.

the machine then becomes what

is

known

as

an induction

generator.

'From the foregoing it is seen that an induction generator is nothing more nor less than an induction motor with some source of motive power applied to its rotor to drive it faster than the stator 's magnetic field. An inductor motor used to drive an elevator acts as in induction generator when the car traveling in the down motion under heavy load the motor

is

;

becomes a generator and pumps current back into the system, which causes it to act to prevent the car from racing.

ALTERNATING-CURRENT MOTORS

401

Induction generators have been used to some extent, driven by low-pressure turbines in parallel with reciprocatingengine-driven alternators, the steam turbine being run without a governor and the speed being taken care of by the steamengine governor. Probably the most notable installation of this kind is that in the Interborough Rapid Transit Company's 59th Street plant in New York City, made a number

In this plant there are five of these machines, of years ago. each of 7,500-kw. capacity, driven by low-pressure steam turbines operating in parallel with the 7,500-kw. steam-enginedriven units. Other cases where induction generators are used are in small water-power plants tied in on a large system. The waterwheels are run without a governor, consequently the machine operates up to its full capacity continuously. Such plants are generally operated without an attendant,

except probably an inspection once per day.

The

chief advantages of such machines

struction

and the

current excitation.

fact

is

their sturdy con-

that they require no source of

On

capable of exciting their

direct-

the other hand, the machines are not

own

field,

consequently must operate

If it were posin parallel with standard-type alternators. sible to maintain a leading power factor on the system, an

induction generator could be operated alone after it had been brought up to speed and connected in on the line. However, this

has not been found feasible in general practice.

CHAPTER XXVIII STARTING POLYPHASE MOTORS Wound-rotor Induction Motors.

At

the instant of start-

ing, the simple squirrel-cage induction motor acts very much like a potential transformer with the secondary short-circuited.

When

the rotor

conductors

is

at rest the electromotive force

induced in

maximum and

the frequency is the same as the stator winding, consequently, since the resistance of the rotor is low the current will be large at a low power factor, its

is

at a

causing the stator winding to take a large current from the with a correspondingly low power factor.

line

The starting torque of an induction motor is influenced to a very large degree by the angle of lag between the current and the e.m.f. in the rotor, being at a maximum when the two are in phase and at zero when the current is lagging 90 deg. Since the phase relation between the current and the e.m.f. in an alternating-current circuit depends upon the resistance and the reactance of the circuit, the two being nearly in step when the reactance is small and the resistance high and differ-

ing in phase nearly 90 deg. when the reactance is high and the is low, it is evident that to provide a good starting

resistance

torque the resistance of the rotor should be high and the reactance low: however, high-resistance rotors decrease the efficiency of the motor when running, therefore, to get the best operating results the rotor should have high resistance at starting and low resistance when running. This condition is

frequently secured by the use of a wound rotor connected to resistance, a motor of this type permits easy

an external

speed adjustment as will be explained later. This type of rotor instead of being wound with bars

which are short-circuited at each end as in the squirrel-cage 402

STARTING POLYPHASE MOTORS

403

type, has a winding of insulated wire similar to the winding used on the armature of a three-phase alternator, the polar

spacing depending upon the polar spacing of the stator windThis winding is connected in series with an external ing. resistance at starting which to speed. It is

is cut out as the rotor comes up 449 and 450, Chapter XXVII.) (See Figs. well known that the torque of a wound-rotor motor

depending upon the position of the rotor winding with respect to that of the stator. This fluctuation increases as the current increases and the smaller the number fluctuates greatly,

For

of phases in the rotor.

this latter reason the three-phase

winding is used exclusively on the rotors of all polyphase wound-rotor induction motors whether the motor is built for a two-phase or a three-phase circuit. This fluctuation is also present in the squirrel-cage motor, but to a very small degree as the number of phases in the rotor is very great.

Wound

Types of Rotors.

rotors are built in two gen-

and shortmounted within the rotor, and those that have the starting resistance and cutout switch mounted outside

eral types, those that have the starting resistance

circuiting switch

the motor.

FiG. 468.

Phase-wound rotor with internal starting

Fig. 468 shows a

mounted on the

wound

spider.

The

resistance.

rotor in which the resistance resistance

is

made

is

in three parts,

one part for each phase, and consists of cast-iron grids inclosed in a triangular frame which is bolted to the end plates

ELECTRICAL MACHINERY

404

holding the rotor laminations together. It is short-circuited by sliding laminated spring-metal brushes along the grids. These brushes are supported by a metal sleeve upon the shaft

which is operated by a rod that passes through the end of a hollow shaft and engages the brush arrangement in the rotor. The outer end of this rod terminates in a loose knob as shown,

and by pushing

in the

knob when the motor has attained

its

In the larger-sized motors the to the bearing bracket. secured lever a by forms: Castdifferent in three These resistances are made

speed the resistance resistance is cut out

is

cut out.

iron grids, cast-brass grids

and cylindrical

coils

made from

a

german-silver strip wound on its edge. the resistance is cut out in a way similar to that explained. If the resistance is to be cut out at a distant point from the

In

all

three forms

motor, or the motor is to be used for variable-speed service it is best to mount the resistance and controller external from the motor.

This necessitates the use of three collector rings,

and keyed to the shaft. Fig. 450, Chapter XXVII, shows a rotor of this type. These collector rings connect the rotor winding to the controller and resisttance through brush gear provided for that purpose. Fig. 469 shows diagrammatically the rotor winding connected to the starting resistance through the collector rings and brushes. The three rotor windings A, B and C are connected in star, and the terminals of the windings are connected to the three collector rings. The three branches R^ R 2 and R 3

insulated from each other

of the starting resistance are also connected in star by the three-armed short-circuiting switch S. At starting the arm is

is

in the position shown by the full lines and all the resistance in series with the rotor winding. As the rotor comes up to

speed the arm is gradually moved around in the direction indicated to the position shown by the dotted lines. At this position the resistance is all cut out and the three rotor windings are short-circuited by the arms of the switch. By using three resistances, one in each phase of the rotor, and a short-circuiting switch, as shown in Fig. 469, the resistance in each

phase

is

kept balanced, and, consequently, the current for the

STARTING POLYPHASE MOTORS

405

different starting points, although this is not absolutely necessary except at the point of maximum starting torque.

FIG.

469.

Balanced starting resistance for induction motor.

In most cases satisfactory starting can be obtained by is known as an unbalanced resistance, as shown in

using what Fig. 470.

Two

FIG. 470.

resistances

R

r

and

R

2

are connected in open

Unbalanced starting resistance for induction motor.

delta at starting by the arm full lines, as the motor comes

moved around

8 in the position shown by the up to speed the arm is gradually

to the position indicated

by the dotted

lines, at

ELECTRICAL MACHINERY

406

which position the resistance

is all

cut out and the rotor wind-

ings are short-circuited. With such a connection the current in phase C is greater than in phases and B for all starting the in current each points, phase becoming nearer balanced as

A

is decreased until the rotor windings are shortwhich point the conditions are the same as in Fig. 469. This unbalancing of the current has so little effect that it is unobjectionable for most starting purposes. Wound-rotor Motors that Start Automatically. An interesting type of polyphase induction motor is one in which the rotor has a high resistance at starting and a low resist-

the resistance circuited, at

ance when running. An insulated winding similar to that used on the armature of a direct-current motor is placed in the bottom of the rotor slots and connected to a commutator. As there

is

no external circuit provided for

rent will be set

up

to a squirrel-cage top of the slots.

in, it

at starting.

A

this

winding no cur-

second winding similar

winding of high resistance is placed in the starting this winding acts like that of a regular squirrel-cage motor except on account of the high rotor resistance the starting current taken from the line is considerably reduced, and the starting torque increased. When the rotor comes up to speed a short-circuiting device mounted on the shaft at the outer end of the commutator is thrown in by a centrifugal governor and short-circuits the coils of the winding in the bottom of the slots through the

At

commutator, making this winding similar to the squirrel-cage winding and thus reducing the resistance of the rotor. At running the rotor is similar to one that has two squirrel-cage windings in parallel. The commutator is used for no other purpose than to provide a ready means to short-circuit the winding, thus making the motor automatic in its starting and requiring no attention other than closing the line switch at

and opening it again when the motor is stopped. In another type of wound-rotor induction motor that is automatic in starting, the rotor winding is similar to that used on a direct-current armature and is connected to a vertical

starting

commutator.

This winding

is

so connected that the coils are

STARTING POLYPHASE MOTORS in series at starting

and thus increase the rotor

407 resistance.

After the rotor has attained the proper speed a short-circuiting device mounted within the rotor is thrown in by a centrifugal governor and the rotor is then running as one of the

The stator winding is the same as that squirrel-cage type. used in any polyphase motor. This type of motor is well adapted to remote control, requiring no controlling device other than a single-throw three-pole switch to open and close the line circuit and a double-throw switch if the motor is to be reversed.

To Reverse the Direction

To reverse the

of Rotation.

direction of rotation of a two- or three-phase induction motor it is necessary to reverse the direction of the revolving mag-

In a two-phase motor this can be done by crossing the terminals of either phase with the terminals of the motor, and a three-phase motor can be reversed by crossing any two netic field.

terminals.

The following diagrams

will

make

this clear:

Fig. 471 represents a two-phase induction-motor winding connected to a four-wire, two-phase circuit crossing the con;

A

nection of phase or B, as shown in Fig. 472, will reverse the direction of rotation. Fig. 473 represents a two-phase motor connected to a three-wire, two phase circuit. The voltage

E

the same, but between the two outside legs, motor under this condition can be reversed in

across each phase it is

E \/2-

The

is

first as shown in Fig. 474, by crossing either of the phase terminals in the motor, which leaves the connections with respect to the e.m.f. as they were at first.

two different ways,

is shown in Fig. 475 and consists in crossline the outside terminals. This gives the same voltage ing relation with reference to the two different windings, that is,

The second method

there

is

E

volts across phases

two outside terminals.

A

and

B

and

-EJ-v/2

across the

Care should be taken not to connect a

two-phase motor, as shown in Fig. 476, as this gives an excessive voltage across one phase which changes the phase rela-

two windings and reduces the startthe excessive current is liable to burn out the

tion of the current in the

ing torque, also w'ndings in a very short period.

ELECTRICAL MACHINERY

408

In Fig. 477 is shown diagrammatically the windings of a three-phase motor connected in star to a three-phase circuit. The voltage is the same between any two-line wires, and PhaseT

Phase*"

<

t

Phase~B~

Phase"*"'

-

>,

FIG. 471

Phase"**

Phase"*"

Fia. 472

FiG. 473

FiG. 475

FiG.. 476

Phase'B"

csm <

r

> < ..... r '

FiG.

474

FIGS. 471 to 476.

Diagrams of two-phase motor

circuits.

between any two terminals of the motor there are two windings connected in series. Hence, any two of the line terminals can be crossed with respect to the motor terminals and have the same relation between the line voltage and the windings Phase

FIG.

FIG. 478

477

FIGS. 477 to 479.

in the motor.

B

FIG. 479

Diagrams of three-phase motor

circuits.

This also holds true for the delta connection

shown in Fig. 478 in this connection, however, there is only one winding across each phase and to work on the same voltage as the star connection will require seven-tenths more turns in ;

STARTING POLYPHASE MOTORS each winding than required in star.

Crossing any

when

409

the windings- are connected shown in

two of the line terminals as

Fig. 478 with respect to the motor terminals will reverse the direction of rotation.

Comparison of Squirrel-cage and Wound-rotor Motors. Although the wound-rotor polyphase induction motor has the most satisfactory starting properties of any type, developing about 100 per cent starting torque for 100 per cent full-load starting current, other starting torques being proportional to the current taken until maximum starting torque is obtained,

and can readily be used for variable-speed service. It is, nevertheless, inferior to the squirrel-cage type in every other The cost is higher, the construction not so rugged, respect. the efficiency, power factor, and pull-out torque are lower and the motor itself requires more attention than the squirrelcage type. For these reasons probably 75 per cent of the poly-

phase motors in use to-day are of the squirrel-cage type.

One of the greatest objections to the squirrel-cage motor is the large current at low power factor taken from the line at starting. This heavy starting current is very objectionable for several reasons. If the motor is started on a lighting circuit the heavy starting current causes the lamps to flicker and if the motor is started very often this flickering of the lamps

becomes very objectionable.

The

excessive current taken at

starting may overload the prime mover and generator supplying the power or the transformers from which the motor

supplied, or the starting current may represent a large percentage of the total power transmitted in the feeder. This

is

may

cause an excessive drop in voltage which may affect other from the same feeder. In some cases the

devices supplied

heavy lagging current

may

cause synchronous devices on the

same system to hunt. To reduce the starting current and improve the starting torque of squirrel-cage motors various devices are used such as auto-transformers, rheostats and special starting coils in the stator windings, the latter will be explained later.

the most

common

is

the auto-transformer.

Perhaps

ELECTRICAL MACHINERY

410

In sizes up to 5-hp. induction motors are usually started by connecting them direct to the line. In sizes of 5 hp. and above an auto-transformer or starting rheostat is used. Some prefer the rheostat for starting small motors

and the auto-trans-

former for large ones. How Auto-Transformer Operates. An auto-transformer has but one winding for both primary and secondary. This type of construction reduces the amount of copper used, depending upon the ratio of transformation. Fig. 480 repreLet it be sents an auto-transformer diagrammatically.

(

(

ai

c

i* c

t

STARTING POLYPHASE MOTORS

411

that the current in AC will flow opposite to the current in be in an inverse proportion to the number of turns in each section of the coil, for theoretically the ampere turns (Tp ) in the primary coil must be equal to the ampere- turns (Ts ) in the secondary coil; hence in section

BC and

EC so

will

the inverse proportion

Ts :TP = I P

:/..

Therefore, Is

which

= TpXlp

will flow in the

f

90X50 =

75

~~f$Q~

direction as

am P eres

shown by the arrowheads and comis flowing through the motor a

bines with the primary current so there current of

125 amperes,

although the generator is only required to supply 50 amperes. Comparison of Auto-Transformer and Resistance for Starting. If a resistance was used to reduce the voltage instead of an auto-transformer, for the motor to have the same starting torque it would have to receive 125 amperes from the line at 100 volts impressed at the motor terminals.

To

get this condition a resistance that will cause a drop of 150 volts with a current density of 125 amperes must be connected in series with This resistance R would equal the voltage drop divided by the motor.

the current, or

#=

150

= 1.2

ohms.

125

This condition is shown in Fig. 481, and the generator has to supply 125 amperes to the motor instead of 50 amperes as when an auto-

I- 125 Amp.

R-/.20/WT7S

f "FiG.

481.

Diagram of starting

transformer was used.

resistance

and motor

circuit.

In the case of the resistance the watts loss equals by the circuit, or

the voltage drop across the resistance multiplied

150 X 125

The

total

ance

is

= 18,750

watts.

power supplied by the generator when starting with a

#7=250X125 = 31,250

watts,

resist-

ELECTRICAL MACHINERY

412

the difference between this and the watts lost in the resistance being the power supplied to the motor, or

31,250-18,750 = 12,500 watts.

With the auto-transformer

all the energy supplied by the generator (except a small loss in the coils of the auto-transformer) was used in starting the motor, that is,

El = 250 X 50 = 12,500

watts.

From the foregoing it would seem that an auto-transformer has much superior starting properties to that of a and

would be true if the power factor at starting However, the power factor is very low, and as the drop in the line is largely due to the wattless component of the current and not the power component, with either type of rheostat,

this

were unity.

component is about the same. Therefore would be about the same in one case as in The power component produces some drop and this

starter the wattless

the line disturbance

the other.

greater with the resistance type of starter, but is again offset by the lagging current required to set up the flux in the transis

former

core.

One

feature that any starting device should possess to be entirely satisfactory is automatic adjustment of the voltage at the motor as it comes up to speed, and in this feature the resistance type of starter is superior. At the instant of starting the motor takes a maximum current from the line; this

maximum drop across the starting resistance, and as the motor speeds up it generates a counter e.m.f. which increases with the speed and the current decreases. As the curcauses a

rent decreases the voltage drop across the resistance decreases This will be made across the motor terminals.

and increases

by again referring to Fig. 481. If the condition at the instant of starting is as shown with a starting current of 125 amperes, the drop across the starting resistance is 150 volts clear

with 100 volts impressed on the motor terminals. If after the motor has come up to speed the current decreases to 50 amperes, the voltage drop (E d ) across the resistance will equal the resistance (R) multiplied by the current (7) or

E d = RI =

1.2

X

50

=

60 volts

STARTING POLYPHASE MOTORS

413

The voltage impressed upon the motor terminals will equal the difference between the line voltage (E) and the drop across the resistance, or

Ed =

E Therefore,

the

voltage

250

60

= 190 volts

has automatically increased at the this is not possible with an

motor terminals from 100 to 190

;

auto-transformer, for the secondary voltage is fixed by the ratio of the primary turns to the secondary turns. Moreover, the resistance type of starter can be readily constructed with.

a

number

of intermediate steps.

This means that the voltage

can be gradually raised at the motor and when the latter is thrown directly one the line the voltage change will not be so great, consequently, the current taken from the line will not be so great with

corresponding drop in line voltage. is much cheaper than the autofor its use with small motors. is one reason which transformer, As regards the power taken from the line at starting the

The

its

resistance

auto-transformer

is

starter

distinctly superior to the resistance type

;

hence a large percentage of squirrel-cage motors are started by this means, the resistance type being limited usually to sizes below 25 hp.

FIG. 482.

starting compensator having three auto-transformers.

Diagram of three-phase

Compensators for Polyphase Motors. For starting two-phase motors two single-phase auto-transformers are used, one connected across each phase, and for starting Starting

ELECTRICAL MACHINERY

414

three-phase motors two types of compensators are used, one using a three-phase auto-transformer with the three coils con-

nected in star.

Fig. 482 shows such a starting compensator

At starting the switch is and motor diagrammatically. thrown to the position shown by the dotted lines. This connects the motor through the auto-transformer to the line beyond the fuses, which would be blown by the heavy starting current if left in the circuit. Assuming an instant when the line polarity is as indicated, that is, A and C are positive and

B

negative, then

when

the switch

is

closed, current will flow

from A and C through sections D and F of the auto-transformer through the motor winding and back through section E of the auto-transformer to the middle terminal of the line B, This current flowing through secas shown by the arrows. F tions D, E and of the auto-transformer induces an e.m.f. in sections d, e, and /, opposite to the e.m.f. in sections D, E and F ; therefore, a secondary current, having the direction shown will combine with the primary current and flow through the motor. This is identical to the conditions represented in Fig. 480, only in this case three transformers are used instead of one. After the motor comes up to speed the switch is thrown

over in the opposite position and the motor through the fuses.

is

connected direct

to the line

FIG.

483.

Diagram of three-phase starting compensator having two auto-transformers.

The second type to start three-phase

of starting compensator that can be used is one that employs two auto-trans-

motors

STARTING POLYPHASE MOTORS

415

formers connected in open delta. This type has the advantage that it can be used for starting either three-phase or two-phase motors. Fig. 483 shows such a starting compensator connected to a three-phase

motor, the action being similar to that of

Fig. 482.

mini

i

FIG.

484.

m

Three-phase starting compensator with cover removed.

Fig. 484 shows a compensator with the cover removed to show the various parts in the center are the three coils of the ;

ELECTRICAL MACHINERY

416

D

The double-throw switch three-phase auto-transformers. operated by the handle E, which is interlocking and cannot be thrown to the running position until first thrown to the

is

F

and G protect the motor starting position. Overload relays from overloads and may be replaced by fuses; is a no-

H

voltage release coil which,

when

the line voltage

D

fails, releases

E

the starting switch and handle and they come back to the off position so that the motor will not be started when the

power comes back on the line, except by an attendant. The oil tank in which the switch D is immersed is shown at the bottom.

A a

motor that has to start under a heavy load will require voltage than one starting under light load and

much higher

taps are usually brought out so that about 35, 50, 65 or 85 per cent of the line voltage may be impressed on the motor terminals. With some starters, employing a drum switch im-

mersed in

oil,

two or more voltage taps are brought into action

in steps.

Rheostats for Starting Polyphase Motors. are used for starting three-phase

rheostats

motors

the balanced and

Two

types of

squirrel-cage the unbalanced types; the former

employs three resistances, one connected in each leg of the circuit, and usually arranged so that resistance can be cut out in equal steps, as shown diagrammatically in Fig. 485. The connections are

made

so that the fuses are out of circuit as

with the auto-transformer at starting. For starting, the switch is thrown to position ( 1 ) which connects the motor to the line with all the resistance in series; as the motor speeds ,

is thrown to position (2), this cuts out one section of the resistance in each leg of the circuit, causing a further increase in speed. The switch is next thrown to

up the switch

position

(3),

which connects the motor direct

to

the line

through the fuses. Contacts (1) and (2) are arranged so that they are out of circuit when the switch is in the running position.

The unbalanced type has two resistances, one in each of le^s of the circuit, as shown diagrammaticaJly in Fig. 486.

two

STARTING POLYPHASE MOTORS

417

These are arranged so that they can be cut out of the circuit similar to Fig. 485. Although this type costs less than the balanced type the starting current is much larger for a given

starting torque and cannot be recommended except in cases where cost is of primary importance. The curves in Fig. 487 show the percentage of full-load starting current required by

a 20-hp. motor in the three different cases in terms of the IP

.5

fuse

Fust

ELECTRICAL MACHINERY

418

by the ratio of the transformer and is, therefore, theoretically the same for a large motor as for a small one. ing

is

fixed

600

FIG. 487.

Per Cent.of Full Load Torque Percentage of full-load current required and correspond torque of an induction motor.

It is general practice to furnish

each motor with

its

own

;

stai

ing device but is sometimes convenient to arrange to stn several motors from the same starting compensator. scheme

A

FIG. 488.

Connection for starting a number of induction motors from the same compensator.

shown in Fig. 488. In addition to the starting compensator, which must be of a capacity equal to that of the largest motor started by it, a three-pole double-throw for doing this

is

STARTING POLYPHASE MOTORS

419

switch must be provided for each, motor. When the switch is thrown to the lower position the motor is connected to the line

through the starting compensators, alter

it

has come up to

speed the three-pole switch is thrown to the up position, thus The auto-transconnecting the motor directly to the line. former is then brought back to the off position ready to start

up any other motor

desired.

Fuse

FIG. 489.

Showing additional turns

in stator

winding for starting.

A

squirrel-cage motor may be provided with additional turns in the stator windings for starting, as shown diagram-

matically in Fig. 489. When the three-pole switch is thrown to the starting position all the turns in the windings are in series after the motor has come up to speed, the switch is then ;

thrown over winding

is

stator core

to the

cut out.

running position and part of the stator

To provide room

and frame have

to be

for the starting coils the

made

larger than otherwise.

ELECTRICAL MACHINERY

420

Another method used in starting three-phase squirrel-cage is the star-delta method used for starting motors of

motors

than 30 hp. At starting, the three stator windings are connected in star, as shown in Fig. 490. The voltage across each winding is equal to the line voltage (E) divided by the -v/3 or about 57 per cent of the line voltage: this is usually less

motor under most conditions. When the motor comes up to speed the switch is thrown to the running position, and the windings are connected in delta, as in Fig.

sufficient to start the

491

;

this applies full-line voltage across each winding.

FIG. 490

It

FIG.

490.

FIG.

491.

Three-phase star-connected winding. Three-phase delta-connected winding.

should be mentioned that in some cases

it is

possible to

bring out low-voltage taps from the neutral of the transformers supplying the power and by means of a three-pole

double-throw switch 50 per cent line voltage can be obtained for starting. This requires two sets of wires being run from the transformers to the starting switch at the motor. If the

motor

is

some distance away from the transformers the

the additional conductors starting device.

may

cost of

be equal to or exceed that of a

INDEX Alternating-current, polyphase sys-

tems, 333

Adjustable-speed motors, 232 Alternating voltages and addition

of,

power of, 374 power factor, 363, 371

currents,

326

pump

Alternating current, 175, 307 addition of, by vector diagrams,

addition

of

grams, 339 alternating electromotive

diagrams

voltage or current curves of, 388

watt-hour

of,

Alternating-current generators, 144,

182,282

readings on, 324 current lags behind voltage, 366

adjusting load on, 353 cycles, poles

316

cycles, poles

and speed,

delta

relation

electrical degrees,

325 315

relation

star connections, 344

magnitude of

317

frequency, 321, 352 fundamental principles heating effects, 325

and

fundamental principle 311 in phase, 354

star connections, 344

effective value of,

and speed,

between, 320

between, 320

and

meter connected on

single-phase circuit, 370 wattmeter reads true watts, 369

voltmeter and ammeter

cycle, 315,

347

two-phase systems, 333 four-wire, 335, 337 three-wire, 334, 338

force,

average value of, 325 capacity, hydraulic analogy 372 capacity, effects of, 371 circuit, power in, 370

of,

voltage or current curves, 342

dia-

312

of,

318

alternators,

two-phase voltages

and currents by vector

flow

368

three-phase systems, 339, 343

326

delta

of,

single-phase system, 318, 333

327 addition of voltages and currents,

circuits,

analogy

sine curve, 324

e.m.f.,

of,

307,

311

operating in parallel, 350 of,

putting into service, 354 single-coil type, 144, 166

307

single-phase, 318

kilovolt-amperes, 363, 370

synchronizing connections for, 354 three-phase delta-connected wind-

kilowatts, 363

maximum

value of, 325, 326 or electromotive, 146, 311

ing,

354

star connected windings, 346

421

INDEX

422

Alternating-current generator, three-

phase machine armature, 343 two-phase type, 335

with

ring

four-pole winding, 336

voltage curve, 147

motors Alternating-current Induction Motors), 386

(see

squirrel-cage type, 386

three-phase, 407

two-phase, 407 wound-rotor type, 386, 403 Alternators,

parallel

operating

of,

350 alternators, 353 for,

electrodynamometer 360 Lincoln type, 358 synchromism, 355 synchronizing, 355 connnections

methods

of.

for, 354,

type

of,

of slots

spanned by a

coil,

cosines table

of,

324

and

bismuth

Apparent watts, 369 Arc lamps, 55 Armatures, 152 coils, 159 of,

166

of,

152

Balancer

sets, 277, 285 balanced load on, 279 connections for compound

sets,

voltage balanced,

of voltage control, 280

shunt type, 277 theory of operation, 278 Batteries, 29 ampere capacity, 40 bells in parallel connected

to,

89

charging, 91 current of, how increased, 40 current sources in parallel, pump

developed, 314

couples, 35

types

Asbestos, 33

methods

313 sine of, 307 sine curve,

objections to, 153 slotted core, 159

277

357

and

resistance of, 217

how maintains 356

cosine of, 308

natural sines

leads

number

281, 282 for three-wire systems, 276

Ampere, 32 Angle, 307

Antimony

drum-type, 226, 152 smooth core, 155 objections to, 158 magnetic leakage, 154 eddy currents, 156 effects of, 157 eliminating. 158 laminated cores, 158 large-size, 160

ring-type, 152, 226, 239

350 diagram of two, 352 in phase, 354 putting an alternator into service, 354 synchronoscope, 359 connections for, 361

how

153

of,

152

adjusting load on two or more condition

Armatures, coils, number conductors in, 229 core slots, 160

thermc

analogy

of,

42

effects of increasing the load on,

88 cell, 29 ohmic resistance

galvanic

of,

91

parallel or multiple connection, 40 parallel-series connections, 43, 46

INDEX Batteries, resistance in scries with,

423

Commutators, expansion and contraction, 167

91 series connected,

function

84

45

series parallel connected, 43,

short-circuited, 56

165

large-size, 166,

temperature analogy voltage of, 38

of, 30,

31

169

neutral point on, 174 small-size, 167

how

substitutes for mica, 167 types of bars, 166

to increase, 37 voltage sources in series, 39 voltaic cell, 29 voltage,

of,

insulation of, 167, 168

Compensated type motor, 245

connected to, 86 Bells connected in parallel, 89 Bismuth and antimony thermobell

couples, 35 British thermal unit, 77

compensating windings, 246 circuits, 82 Compound generator, 208 ampere-turns, adjusting, 214 compounding shunt, 213, 214

Complex

density

of

magnetic

circuit,

209 Centimeter-gram-second, 137

Chemical

effects of electric current,

design of series winding, 213 equalizer connection, 293

flat-compounded, 211 210

6 Circular mil, 117

full-load current of,

Commutation, 170 Commutation period, 179 Commutating-pole machines

large-sized machine, 212 long-shunt connection, 214 (see

Interpoles), 241

compound-wound

parallel, operating in,

interpole

ma-

chine, 243

connection

diagram

com-

for

pound-interpole motor, 244 connection diagram for seriesinterpole motor, 244

connection

diagram

for

shunt-

number

used, 241

242 shunt and compound, 242 shunt-interpole type, 242 Commutators, 166 common type of, 167 commutation, 170 brushes shifted to improve, 180 coils under, 176 effects of self-induction on, 178 sparkless, 180, 246 construction of, 166 series-interpole type, series,

292

proper amount of compounding, 212 putting into service, 297 reversed polarity, 295 series

to

interpole motor, 244 interpoles,

overcompounded, 292

winding of, 209, 213 compensate for volts drop,

210. series-field

coil,

large-capacity,

213 strength of, 294 short-shunt connection, 214

series-field,

taking out of service, 299 under-compounded, 213 voltage curve of, 208 voltage at terminals of, 211

motors, 237 connecting up, 261 cumulative compound, 238

Compound

determining series and shunt 262

field,

INDEX

424 Compound

motors, diagram of con-

Delta and star connections, 344 Direct-current generators, 133

nections, 262, 265 differential

compound, 238

elevator service, 237 direction of

reversing

rotation,

266

and

series-

shunt-field

coils

of

armature conductors, 229 armature of, 142 armature reaction, 238 brushes

177

of,

on neutral

opposite polarity, 264 speed variation of, 238

positions

224

testing polarity of series and shunt-field windings, 263

coils

torque characteristic, 237 shunt, 213, 214

commutator, 148

Conductors, 117 of,

carrying capacities circular mil, 117

117, 125 of,

crossing field connections of

122

pound-wound

com-

generator, 201

current to excite field coils

of,

formula, 128

determining direction of electromotive force, 135, 140

cross-section of, 270

economical size of, 123 National Board of Fire Underwriters' requirement, 125 rubber-covered, 122 stranded conductors, 121 varnished-cotton-cloth

covered,

122

diagrams

of,

195,

320

dynamo, 52 Edison two-pole machine, 161 204 efficiency, per cent, 220 effects of loading, 195,

electromagnetic induction, 133 exciting field coils of, 182

voltage drop allowable, 122 voltage regulation of, 123

external characteristic curves

weather-proof or a fire-proof insulation, 122 wire gage, 117

field poles,

wire

sizes,

118

worked out from No. 10

wire,

121

Cycle, 315, 316

159

separately excited, 206

flat-compounded, 211 fundamental principles

how how how

of,

133

loaded, 141

voltage

voltage

is

built up, 183

is

maintained con-

increasing excitation of, 292 large

compound-wound, 201

load on series generator, 193 load on a shunt generator, 193

defined, 51

223, 231

fields

stant, 175, 196

Conductance, 65 Consequent-pole machines, 165 Copper losses in armature, 216 Cosine of angle, 308 Counter-electromotive

of,

208, 291

wire table, 119 features of, 118

Coulomb

216

density of magnetic circuit, 209

128

foot,

under commutation, 176 commercial types of, 165

compounding shunt, 214 compound- wound type, 199

Compounding

calculating size

222

axis,

of,

long-shunt connections, 214 force,

140,

losses in, 215, 219 magnetic flux distribution in,

magnetic leakage, 154

240

INDEX Direct-current

tude of

magni-

generators,

311

c.m.f., 143,

multipole machines, 151, 164

291

equalizer connection, 293, 296 putting generators into service,

297 reasons for operating inparallel,

290 taking generator out

of-

service,

299

compound type, 237 compound-wound interpole

residual

magnetism, 189, 205 rotation

commutating-pole machines, 241 counter-electromotive force, 140, 229, 231

fundamental 222

principle

138,

of,

in,

227

operates when loaded, 229 left-hand rule for a, 140, 228

shunt and compound types, 242

series,

183,

187,

series

of

type,

243

how

principles of, 311 putting into service, 297

armature,

188

type

of,

233

series-interpole type,

242

shunt type, 235

right-hand rule for e.m.f ring- type armature, 148

.

of,

228

generator,

184,

types of armatures, 152 two-pole machines, 151

compound

voltage, contol of, 194

varying the voltage of, 145 volts drop in armature, 206 Direct-current motor, 138 adjustable-speed type, 232

another theory

of,

224

adjusted, 230, 231

190,

206 series, shunt and compound types, 242 shifting the brushes on, 202 short-shunt connection, 214 shunt-field current of, 214 shunt- type, 190 single-coil type, 146, 166 symbol to represent, 52 taking out of service, 299 types of, 182

voltage at terminals of generator, 211

shunt-interpole type, 242 speed, how Dynamo, 52

salient-pole machine, 163

series-type

motor, armature 238 brushes positions of 224 on neutral axis, 222 in,

generator action present

three-wire generators, 301

reversing

Direct-current

reaction

parallel operation of, 290 effects of generator characteristics on,

425

E Eddy

currents in armature core, 156

Eddy-current losses, 217 Efficiency, per cent, 220 Electrical degrees, 315 Electrical horsepower, 79 Electrical instruments, 93 ammeter, 96 connected as an ammeter, 99 to measure volts and amperes, 100 a millivoltmeter, 99 direct-current type, 93

electrodynamometer

type,

104,

323for

use

on

circuits,

alternating-current

104

galvanometer, 94, 112 principle of, 94 inclined-coil type,

managnin, 100

323

INDEX

426 Electrical

instruments, managnin, shunts constructed of, 100

measuring

a

Electricity, electrical resistance, 32 electrical resistance, what con-

or

direct-current

voltage, 94

milliammeters, 99

heat analogy

99

insulator

millivoltrrieters,

unit

force,

137

milliampere, 99 millivolt,

56

stitutes,

electromotive

99

of,

of,

2

34

analogy of, 2 magnetic effect of, 7 light

movable element of, 96 movement of modern type, 95

potential difference

power-factor indicator, 373

sound analogy

principles of, 93 scale range, 97

unit resistance, the ohm, 34

spiral springs of,

of,

29

of, 1

unit of electric current, 32

95

voltage, production

synchronoscope, 359

electrodynamometer type of, 360 Lincoln type of, 358 terminal marking, 95 voltammeter 101 voltmeter, 96 across armature terminals, 204 /

of,

29

Electrodynamometer, 360 Electromotive force, 314 Electroplating, 32 Elevator service, 237 Energy, work and power, 73 Equalizer connection, 293, 296

voltmeter and ammeter reading

on

alternating-current

cir-

324 volts at armature terminals, 204 volts drop in armature, 206 wattmeter, direct reading, 102 cuits,

reads true watts, 369 watt-hour meter, 104 connected on single-phase cuit, 370 dial, 105

cir-

Wheatstone bridge, 111 Electrical pressure, 29 Electrical resistance, 32 Electricity, 1

ampere, 32 chemical effect

of,

6

coulomb defined, 51

Faraday's

first

dynamo, 134

Field magnets, 160

ampere-turns on, 191 160 field coils, 203 function of, 160 horseshoe shape, 161 laminated poleshoes, 165 materials used in, 164 earlier types of,

multipole designs, 164 permanent-magnet, 161 salient-pole machine, 163 Flat-compounded, 211 Foot-pound, 74 Foot-pound-minute, 75 Frequency, 321, 352 Friction losses, 215

coulomb, the unit quantity, 50

G

direction of, 8 electrical pressure, electrical

ture,

29

pressure and tempera-

30

Galvanic

cell,

29

Galvanometer, 94, 112 German silver, 57

of,

INDEX

H

Induction motors, starting ance for, 411

Harmonic motion, 308 Heat conductivity, 33 Heat

427 resist-

two or more motors from same compensator, 417

starting

equivalent of electric current,

synchronous, speed of, 397 synchronous, type of, 321 three-phase, 407

78

Horsepower denned, 75 Hysteresis losses, 217

three-phase delta-connected winding,

420

Inductance, 170 Induction, 5

three-phase star-connected wind-

Induction generator, 397 Induction motors, 386

three-phase

ing,

tors,

additional turns in stator wind-

how

operates,

410 auto-transformer and resistance for starting,

comparison

of,

411 for,

comparison of squirrel-cage and wound-rotor types, 409 diagrams of three-phase, 408 diagrams of two-phase, 408 how rotation is produced, 395 parts of, 386 per cent slip, 397 phase-wound rotor, 387 phase-wound rotor with internal starting resistance, 403 polyphase type, starting, 402 relation between poles, frequency and speed, 393 reverse direction of rotation, 407 revolving magnetic field, 388 rheostats for starting polyphase type, 416 slip,

397

squirrel-cage rotor, 387 squirrel-cage type, 386 star-delta method of starting, 420

stator

of,

compensa-

two-phase, 389, 407

unbalanced starting for, 405 wound-rotor, 402 start automatically,

type

balanced starting resistance 405

starting

415

types of rotors, 403

ings for starting, 419

auto-transformer,

420

386

starting compensators for, 413

of,

resistance

406

386

Internal resistances, 83

Interpole machines, 241

compound-wound connection

type, 243

diagram

com-

for

pound-interpole motor, 244 connection diagrams for series

244

interpoles,

connection

diagram

for

series

interpole motor, 244

connecting up interpole machine, 266 interpoles,

number

used, 241

proper polarity of interpoles, 267 series type, 242

shunt and compound, 242 shunt type, 242 test for polarity, 267 wrong polarity, 268 series,

J Joint resistance, 64

K Kilovolt-amperes, 363, 370 Kilowatts, 78, 220

Kilowatt-hours, 104

INDEX

428

Magnets,

Laminated

160

core,

Lenz's Law, 365

Long-shunt connection, 214 Losses in electrical machinery, 215 copper losses in armature, 216 current to excite the field

coils,

magnetic field, strength of, 16 magnetic force acts like elastic bands, 138 magnetic force, lines of, 136 magnetism residual, 183, 187

magnetism, theory

216

eddy current losses, 217

natural, 10

218 hysteresis losses, 217 shunt-field winding, 216 total losses in, 219 PR losses, 216, 217

permanent, 12

friction losses,

properties

voltmeter-and-ammeter method, 107 with Wheatstone bridge, 112

pieces, 19

14

electromagnets, 22, 27, 238 direction of current and magnetic field relation between,

22

around a number ductors, 25

field

field builds up,

magnetic

field

of con-

how, 24 about a conduc-

tor, 170 determining direction direction of, 22

of,

24

properties of, 27 residual magnetism, 183, 187

right-hand rule for telling polarity of, 23 solenoid, field of,

27

15

horseshoe, 12, 138

how demagnetized, 21 iron, why magnetized, like poles repel, lines of force, 16

17

12

high resistance, 108 known-resistance method, 107

10

bar, 12

a,

of, 10,

unit magnetic pole, 17 Managnin, 100 Measuring insulation resistance, 110 Measuring resistance, 107

attract pieces of iron, how, 18

earth

19

repulsion between like poles, 137 temporary, 12

Magnetic leakage, 154 Magnets, 10

broken into several

of,

poles of a, 14

M artificial,

distribu-

field

magnetic

tion of, 15

Wheatstone bridge, 111 hydraulic analogy of

a,

112

Measurement of current flow, 37 pressure, 36 Mechanical equivalent of heat, 76 form of energy, 215 pressure, 49 Mho, 65 Micanite, 167 Michael Faraday, 133 Milliammeter, 99 Milliampere, 99 Millivolt, 99 Millivoltmeter, 99 Molecules, 217 arrangement of, 218 in armature core, 218 of iron, 19

19

Multipole machines, 151, 164 Mutual inductance, 171, 173

National

Board

of

Fire

Under-

writers' requirements, 125

INDEX

N

Power

429 factor, 371

indicator, 373

Neutral point shifted, 240

Pulsating current, 147

O

R

Ohm's

law, 49 ampere, unit flow, 50 complex circuits, 82 conductance, 65

coulomb

Residual magnetism, 183, 187, 206 Resistance coils, 57

defined, 51

unit quantity of electricity, 50 current, factors that influence flow electrical

what con-

56

remember, 51 internal resistance, 83 to

effects of,

load-voltage curve, 207

82

voltage generated voltage

formula, 65 parallel circuits,

62

problems

in,

relation

nected of, 63,

67

electric circuit,

78

hydraulic analogy

an

in,

206

decreases, 207

between

volts,

to,

con-

206

Series motor, 233

54

amperes

and ohms, 51 series circuits,

of,

voltage curve, 206 voltmeter and ammeter

mho, 65

of

173

Series circuits, 57 Series generator, 206

external characteristic curves, 208

joint resistance, 64

power

Saturation, magnetic, 203 Self -inductance,

resistance,

stitutes,

how

53

of,

Salient-pole machines, 163

57

armature and 233

field

connection diagrams 254

windings of,

of,

233, 252,

connected to starting box, 253

three-electric units, 49 total resistance of any circuit, 63

distinction

between

series

and

unit electromotive force, 137

shunt-types, 235 reverse direction of rotation, 255

voltage drop along a circuit, 60,

starting box, 234

69

testing a, 234

volts across section of resistance,

60 volts,

amperes and ohms, 51

Shunt generator, 203

volts drop in a source of electro-

motive

force,

torque of, varies, 234 Short-shunt connection, 214

85

volts drop through resistance of

armature, 87

connected to load, 220 effect of the load on, 204 external characteristic curves 208, 291

volts unit of pressure, 49

fields

separately excited, 206

operating in parallel, 292 plotting load current, 205 Parallel circuits, 62 Parallel-series connections, 43, 46

Polyphase

circuits,

power

of,

374

putting into service, 297 residual magnetism, 205 saturation, 203

of,

INDEX

430 Shunt generator, taking out of

ser-

299

vice,

used as a motor, 224 voltage curves of, 203

diagrams

parallel operation of, 301

errors in connecting up,

258 for, 235, 252, 255 direction of rotation, 260

between

series

two

and

shunt-types, 235

235 mistake in making connections, 256 speed variation, 236 torque of, 236 Sine of an angle, 307 Sine curve, 324 field coils of,

Sines and cosines, natural, table

of,

313 Slip,

current distribution

in,

275

current in neutral, 276

two

from

developed

two-wire

systems, 273

Edison, 349

how

voltage

is

maintained bal-

anced, 277 neutral of, 271 for

direct-current

cir-

269

Torque characteristics of motors, 237

system, 333

Transformers, 376

Solenoid, 27 1

auto-transformer,

Starting rheostats, 233, 247 action taking place when motor started,

winding divided into 302

Three-wire systems, 269, 271 balancer sets for, 276

cuits,

Single-phase alternator, 318

on,

sections,

voltages

per cent, 397

Sound,

power output of, 287 series winding on, 303 series and interpole windings 304 series-field

connections distinction

284

of,

distribution of current in, 288

volts drop in armature, 206 Shunt-motors, 235 checking connections, 258

common

balanced

Three-wire generators, load on, 283

is

247

manual type of, 233 shunt motor connected to

how

operates,

410 construction of core, 381 core-type, 380

four-

terminal starting box, 252 starting boxes, 250 starting current curve, 250 starting resistance, 247

Synchronizing, 355

Synchronoscope, 359 Synchronous motor, 321

core-type and shell-type, 379 current transformers, 384

diagram of potential, 382 of, 384 potential and current, 376 potential, operation of, 376 principles of operation, 377 grouping

relation of primary volts,

381

shell-type,

380

and secondary

types of potential, 376

Thermocouple, bismuth and antimony, 35 Three-phase system, 339 Three-wire generators, 282

True watts, 369 Two-phase alternators, 335 motors, 407 systems, 333

INDEX

U

431

Uniform motion, 308

Voltage curve, 147 Voltaic cell, 29

Useful watts, 369

Voltmeter, 96

Unit of power defined, 74

W Watts, 78 Watt-hours, 104

Voltammeter, 101 Volt-amperes, 369

Wheatstone bridge, 111 Wire table, 119

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